Math 1 (Calculus) - First Semester Lectures - Tikrit University

MATH 1 (CALCULUS) FIRST SEMESTER FACULTIY OF COMPUTERS AND IN- FORMATION  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Prelim...

MATH 1 (CALCULUS) FIRST SEMESTER FACULTIY OF COMPUTERS AND IN- FORMATION  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department CHAPTER 1 PRELIMINARIES Real Numbers and the Real Line 1.1 Calculus is based on the real number system. Real numbers are numbers that can be. expressed as decimals :We distinguish three special subsets of real numbers …,The natural numbers, namely 1, 2, 3, 4.1 …,The integers, namely 0, ±1, ±2, ±3.2 The rational numbers, which are ratios of integers. These numbers can be.3 expressed in the form of a function m/n, where m and n are integers and n≠0. : Examples are 200 5 5− 5 1 =167 ,13 =−33= 3− ,2 67 , 3 0 0 0 Recall that division by is always ruled out, so expressions like and are( ).undefined The real numbers can be represented geometrically as points on a number line. called the real line, as in Figure 1.1 Figure 1.1 Intervals 1.1.1 Certain sets (or a subset) of real numbers, called intervals, occur frequently in calculus and correspond geometrically to line segments. For example, if a ˂ b, the open interval from to consists of all numbers between a and b is denoted by the :symbol (a, b). Using set-builder notation, we can write 1 1  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department }x| a ˂ x ˂ b{ = )a, b( ).which is read “(a, b) is the set of x such that x is an integer and a ˂ x ˂ b( Notice that the endpoints of the interval -namely, a and b- are excluded. This is indicated by the round brackets and by the open dots in Table 1.1. The closed interval from a to b is the set }x| a ≤ x ≤ b{ = ]a, b[ Here the endpoints of the interval are included. This is indicated by the square brackets [ ] and by the solid dots in table 1.1. It is also possible to include only one.endpoint in an interval, as shown in Table 1.1 Table 1.1 Inequalities 1.1.2 The process of finding the interval or intervals of numbers that satisfy an inequality. in x is called solving the inequality The following useful rules can be derived from them, where the symbol =˃ means ”.implies“ 2 2  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Example 1: Solve the following inequalities and show their solution sets on the.real line :Solution The solution set is the open interval (-∞, 4) (Figure.1.1a) The solution set is the open interval (-3/7, ∞) (Figure.1.1b) 3 3  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department The inequality 6/(x – 1) ≥ 5 can hold only if x ˃ 1 because otherwise 6/(x – 1) is undefined or negative. Therefore, (x – 1) is positive and the inequality will be preserved if we multiply both sides by (x – 1) and we have The solution set is the half-open interval (1, 11/5] (Figure 1.1c) Figure 1.2 Absolute Value 1.1.3 The absolute value of a number x, denoted by | x |, is the distance from x to 0 on the real number line. Distances are always positive or 0, so we have for every number xx | ≥ 0 | :Or it can be defined by the formula 4 4  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department 𝑥 ≥ 0𝑥, 𝑥− 𝑥 < 0 { =|x| , :Example 2 | ,5 = )5-( - = | 5- | ,0 = | 0 | ,3 = | 3 | | a || = | a | - Geometrically, the absolute value of x is the distance from x to 0 on the real number line. Since distances are always positive or 0, we see that | x | ≥ 0 for every real ,number x, and | x | = 0 if and only if x = 0. Also x – y | = the distance between x and y |.on the real line (Figure 1.2) Figure 1.3 The absolute value has the following :properties 5 5  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department :Example 3 The following statements are all consequences of the definition of absolute value and are often helpful when solving equations or inequalities involving absolute :values The inequality | x | ˂ a says that the distance from x to 0 is less than the positive number a. This means that x must lie between – a and a, as we can see from Figure.1.4 Figure 1.4 Example 4: Solve the equation | 2x – 3 | = 7 :Solution The solutions of | 2x – 3 | = 7 are x = 5 and x = - 2 6 6  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Example 5: Solve the inequality |5 − |< 2 1 The symbol ˂=˃ is often used by mathematicians to denote the “if and only if”( )”. logical relationship. It also means “implies and is implied by The original inequality holds if and only if (1/3) ˂ x ˂ (1/2). The solution set is the.open interval (1/3, 1/2) Lines, Circles, and Parabolas.2 Coordinate Geometry and Lines.1 The points in a plane can be identified with ordered pairs of real numbers. We start by drawing two perpendicular coordinate lines that intersect at the origin O on each line. Usually one line is horizontal with positive direction to the right and is called the x-.axis; the other line is vertical with positive direction upward and is called the y-axis Any point P in the plane can be located by a unique ordered pair of numbers as :follows Draw lines through P perpendicular to the x- and y-axes. These lines intersect the axes in points with coordinates and as shown in Figure 1.5. Then the point P is assigned the ordered pair (a, b). The first number a is called the x-coordinate (or abscissa) of P; the second number b is called the y-coordinate (or ordinate) of P. We say that P is the point with coordinates (a, b), and we denote the point by the symbol P (a, b)..Several points are labeled with their coordinates in Figure 1.6 7 7  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.6 Figure 1.5 This coordinate system is called the rectangular coordinate system or the.Cartesian coordinate system The plane supplied with this coordinate system is called the coordinate plane or.the Cartesian plane The x- and y-axes are called the coordinate axes and divide the Cartesian plane into four quadrants: First quadrant, Second quadrant, Third quadrant and Fourth quadrant as shown in Figure 1.6. Notice that the First quadrant consists of those.points whose x- and y-coordinates are both positive Example 6: Describe and sketch the regions given by the following sets: (c) {(x, y) ||y| ˂ 1}(b) {(x, y) | y = 1}(a) {(x, y) | x ≥ 0} :Solution The points whose x-coordinates are 0 or positive lie on the )a( y-axis or to the right of it as indicated by the shaded region.in Figure 1.7 (a) 8 8  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.7 The set of all points with y-coordinate 1 is a horizontal line one unit above )b(.the x-axis [see Figure 1.7(b)] y| ˂ 1 if and only if -1 ˂ y ˂ 1| )c ( The given region consists of those points in the plane whose y-coordinates lie between -1 and 1. Thus the region consists of all points that lie between (but not on) the horizontal lines y = 1 and y = -1. [These lines are shown as dashed lines in Figure 1.7(c) to indicate that the points on these lines don’t lie in the set.] Increments and Straight Lines 1.2.2 When a particle moves from one point in the plane to another, the net changes in its coordinates are called increments. They are calculated by subtracting the coordinates of the starting point from the coordinates of the ending point. If x :changes from x1 to x2 the increment in x is x = x2 – x1∆ Example 7: In going from the point A(4, -3) to the point B(2, 5) the increments in the x- and y-coordinates are ∆y = 5 – (-3) = 8x = 2 – 4 = -2,∆ From C(5, 6) to D(5, 1) the coordinate increments are ∆y = 1 – 6 = -5x = 5 – 5 = 0,∆ 9 9  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department See Figure 1.8 Figure 1.8 Slope of straight line 1.2.3.Slope is a measure of the steepness of the line Given two points P1 (x1, y1) and P2 (x2, y2) in the plane, we call the increments ∆x = x2 – x1 and ∆y = y2 – y1 the run and the rise, respectively, between P1 and P2. Two such points always determine a unique straight line (usually called simply a line).passing through them both. We call the line P1 P2 Any nonvertical line in the plane has the property that the ratio 𝑟i𝑠𝑒 ∆𝑦 𝑦2−𝑦1 𝑚 = = = 𝑥2−𝑥1∆𝑥 𝑟𝑢𝑛 has the same value for every choice of the two points P1 (x1, y1) and P2 (x2, y2) on the line (Figure 1.9). This is because the ratios of corresponding sides for similar.triangles are equal 1 0 10  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.9 Figure 1.10 shows several lines labeled with their slopes. Notice that lines with positive slope slant upward to the right, whereas lines with negative slope slant downward to the right. Notice also that the horizontal line has slope 0 because.y = 0 and the slop of the vertical line is undefined because ∆x = 0∆ Figure 1.10 1 1 11  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Example 8: find the slop of the nonvertical straight line L1 passes through the. points P1 (0, 5) and P2 (4, 2) and L2 passes P3 (0, -2) and P4 (3, 6) Solution: : Line L1 The slope of L1 is 𝑚𝑥= = ∆𝑦 2−𝑥 1𝑥 ∆ 𝑦2−𝑦1 )2−(−6 3 = 3−0 = 8.That is, y increases 8 units every time x increases 3 units :Line L2 The slope of L2 is 𝑚𝑥= = ∆𝑦 2−𝑥 1𝑥 ∆ 𝑦2−𝑦1 ∆𝑦 2−5 4−0 𝑚 = 𝑥=∆ = 4 −3.That is, y decreases 3 units every time x increases 4 units Lines L1 and L2 explained in Figure 1.11 Figure 1.11 1 2 12  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Equation of straight line.4 Point-Slope Form of the Equation of a Line )a( Now let’s find an equation of the line that passes through a given point P1 (x1, y1) and has slope m. A point P(x, y) with x ≠ x1 lies on this line if and only if the :slope of the line through P1 and P is equal to m; that is 𝑦−𝑦1 =m 𝑥−𝑥1 :This equation can be rewritten in the form y – y1 = m (x – x1) and we observe that this equation is also satisfied when x = x1 and y = y1. Therefore.it is an equation of the given line.Example 9: Find an equation of the line through (1, -7) with slope – 1/2 :Solution Using Point-slope form of the equation of a line with m = -1/2, x1 = 1 and y1= -7, :we obtain an equation of the line as y + 7 = -1/2 (x – 1) :which we can rewrite as x + 2y +13 = 0or2y + 14 = - x +1 Example 10: Write an equation for the line through the point (2, 3) with slope -3/2 :Solution We substitute x1 = 2, y1= 3 and m = -3/2 into the point-slope equation and obtain 1 3 13  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department y = 3 - 3/2 (x -2), or y = - 3/2 (x) + 6.Example 0, yFind When x =11: an the = 6 so equation of the line line intersects thethrough they points y-axis at = 6 (-1, 2) and (3, -.4) 2:By −4 − A Line Through Two Points)b( :Solution 𝑚 − = 2 )1−( − 3 Definition the slope of the line = 3 :Using the point-slope form with x1 = -1 and y1 = 2, we obtain y - 2 = - 3/2 (x + 1) or 3x + 2y = 1.Example 12: Write an equation for the line through (-2, -1) and (3, 4) Solution: The line’s slope is Either way, y = x + 1 is an equation for the line (Figure 1.12) 1 4 14  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.12 Slope-Intercept Form of The Equation of a Line)c(.Suppose a nonvertical line has slope m and y-intercept b. (See Figure 1.13) Figure 1.13 This means it intersects the y-axis at the point (0, b), so the point-slope form of the : equation of the line, with x1 = 0 and y1 = 0, becomes y – b = m (x – 0) :This simplifies as follows 1 5 15  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Example : Find the intercepts of the axis of the equation y = x2 – 1 x2 = 1 → → x2 – 1 = 0 →Solution: For x-intercept, let y = 0 → y = -1 y = 0 -1 → x=±1 For y-intercept, let x = 0 In particular, if a line is horizontal, its slope is m = 0, so its equation is y = b, where b is the y-intercept (see Figure 1.14). A vertical line does not have a slope, but we can write its equation as x = a, where a is the x-intercept, because the x-coordinate. of every point on the line is a Figure 1.14 Example 13: Write the standard equations for the vertical and horizontal lines. through point (2, 3) :Solution )2, 3( = )a, b( x-intercept = a = 2 y-intercept = b = 3 :Horizontal line equation - y=mx+b x +3 0= y =3 :Vertical line equation - x= 2 two lines was shown in Figure 1.15 1 6 16  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.15 The General Equation of a Line)d( :A linear equation or the general equation of a line can be written in the form.where, A, B, and C are constants and A and B are not both 0 :We can show that it is the equation of a line If B = 0, the equation becomes Ax + C = 0 or x = - C/A, which represents a -. vertical line with x-intercept - C/A 𝐶 𝐴 :If B ≠ 0, the equation can be rewritten by solving for y - 𝑦=− 𝑥 𝐵 − We recognize this as being the slope-intercept form of the equation of a line (m= - 𝐵. A/B, b= - C/B) Example 14: Find the slope and y-intercept of the line 8x + 5y = 20 :Solution: Solve the equation for y to put it in slope-intercept form 1 7 17  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department The slope is m = - 8/5. The y-intercept is b = 4 Parallel and Perpendicular Lines)e( Slopes can be used to show that lines are parallel or perpendicular. The following :facts are proved Example 15: Find an equation of the line through the point (5, 2) that is parallel to. the line 4x + 6y +5 = 0 5 2 Solution: The given line can be written in the form 𝑦= − 𝑥 3 − Parallel lines have the same slope, 6 which is in slope-intercept form with m = - 2/3. so the required line has slope – 2/3 and its equation in point-slope form is y – 2 = - 2/3 (x – 6).We can write this equation as 2x + 3y = 16.Example 16: Show that the lines 2x + 3y = 1 and 6x - 4y – 1 = 0 are perpendicular Solution: The equations can be written as 1 8 18  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Distance and Circles in the Plane 1.2.5 To find the distance | P1P2 | between any two points P1 (x1, y1) and P2 (x2, y2), we note that triangle P1P2P3 in Figure 1.16 is a right triangle, and so by the :Pythagorean Theorem we have 𝑃1𝑃2| = √|𝑃1𝑃3|2 + |𝑃2𝑃3|2= √|𝑥2 − 𝑥1|2 + |𝑦2 − 𝑦1|2| 2 + (𝑦2 − 𝑦1)2)𝑥2 − 𝑥1(√= Figure 1.16 1 9 19  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Example 17: The distance between (1, -2) and (5, 3) is Second-Degree Equations In the proceeding sections we saw that a first-degree, or linear, equation Ax + By + C = 0 represents a line. In this section we discuss second-degree equations such as + =1 𝑥2 𝑦 2 y = x2 + 1x42 + 9y2 = 1 x2 - y2 = 1.which represent a circle, a parabola, an ellipse, and a hyperbola, respectively Circles)a( To find an equation of the circle with radius r and center (h, k), by definition, the circle is the set of all points P(x, y) whose distance from the center C(h, k)is r. (See.Figure 1.17) Figure 1.17 :Thus P is on the circle if and only if |PC| = r. From the distance formula, we have 2 + (𝑦 − 𝑘)2 = 𝑟)𝑥 − ℎ(√ or equivalently, squaring both sides, we get 2 0 20  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department 2 + (y – k)2 = r2)x – h( :Example 18 :The standard equation for the circle of radius 2 centered at (3, 4) is )a( 2 + (y – 4)2 = 22= 4)x – 3( The circle )b( 2 + (y +5) =3)x – 1( 2.Has h = 1, k = -5 and r = √3. The center is the point (h, k) = (1, -5) and the radius is r = √3 Example 19: Find the center and radius of the circle.x2 + y2 +4x – 6y -3 =0 Solution: We convert the equation to standard form by completing the squares in x :and y 2 1 21  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department.The center is (-2, 3) and the radius is r = 4 Example 20: Sketch the graph of the equation x2 + y2 +2x – 6y +7 =0 by first. showing that it represents a circle and then finding its center and radius Comparing this equation with the standard equation of a circle, we see that h = -1, k and r = √3, so the given equation represents a circle with center (-1, 3) and radius 3 =.r = √3. It is sketched in Figure 1.18 2 2 22  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.18 Parabola)b( The geometric properties of parabolas will be reviewed later. Here we regard a.parabola as a graph of an equation of the form y = ax2 + bx + c Example 21: Draw the graph of the parabola y = x2 :Solution We set up a table of values, plot points, and join them by a smooth curve to obtain. the graph in Figure 1.19 2 3 23  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.19 Figure 1.20 shows the graphs of several parabolas with equations of the form for various values of the number a. In each case the vertex, the point where the.parabola changes direction, is the origin Figure 1.20 :We see that The parabola y = ax opens upward if a ˃ 0 2 The parabola y = ax opens downward if a ˂ 0. (as in Figure 1.21) 2 2 4 24  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.21 :Note that The graph of an equation is symmetric with respect to the y-axis if the equation is unchanged when x is replaced by – x, and The larger the value of | a | the narrower the parabola ,Generally Example 22: Graph the equation − 2 𝑥 2 − 𝑥 + 1 4 2 5 25  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Solution: Comparing the equation y = ax2 + bx + c with we see that c = 4b = -1,a = - ½, Since a ˂ 0 the parabola opens downward. From Equation (2) the axis is the vertical line 𝑏 )1−( = = 𝑥 =2 − −(22 1− 𝑎 − )1 1 9 When x = -1, we have 𝑦= − (−1)2 − (−1) + 4 = 2 2.The vertex is (-1, 9/2) :The x-intercepts are where y = 0 x2 – x + 4 = 0 )1/2( - x2 + 2x – 8 = 0 (x – 2) (x + 4) = 0 x = 2, x = - 4 We plot some points, sketch the axis, and use the direction of opening to complete the graph in Figure 1.22 Figure 1.22 2 6 26  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department If we interchange x and y in the equation y = ax2, the result is x = ay2, which also represents a parabola. The parabola x = ay2 opens to the right if a ˃ 0 and to the left if a ˂ 0. (See Figure 1.23). This time the parabola is symmetric with respect to the.x-axis because if (x, y) satisfies x = ay2, then so does (x, -y) Figure 1.23 ELLIPSES)c( The curve with equation 𝑦2 𝑥2 + =1 𝑎2 𝑏2 where a and b are positive numbers, is called an ellipse in standard position. :The most important properties of the ellipses are The ellipse is symmetric with respect to both axes, i.e the above Equation is -. unchanged if x is replaced by – x or y is replaced by –y The x-intercepts of a graph are the x-coordinates of the points where the - graph intersects the x-axis. They are found by setting y = 0 in the equation of. the graph 2 7 27  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department The y-intercepts of a graph are the y-coordinates of the points where the - graph intersects the y-axis. They are found by setting x = 0 in the equation of the graph. See Figure 1.24 Figure 1.24 Example 23: Sketch the graph of 9x2 + 16y2 = 144. :Solution: We divide both sides of the equation by 144 𝑥2 𝑦2 1= + 9 16 The equation is now in the standard form for an ellipse, so we have a2 = 16, b2 = 9, a = 4 and b = 3. The x-intercepts are ± 4; the y-intercepts are ± 3. The graph is.sketched in Figure 1.25 2 8 28  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.25 HYPERBOLAS)d( The curve with equation 𝑦2 𝑥2 − =1 𝑎2 𝑏2 where a and b are positive numbers, is called a hyperbola in standard position. :The most important properties of the hyperbola are The hyperbola is symmetric with respect to both axes, i.e the above - Equation.is unchanged if x is replaced by – x or y is replaced by –y The x-intercepts of a graph are the x-coordinates of the points where the - graph intersects the x-axis. They are found by setting y = 0 in the equation of.the graph: y = 0 obtain x2 = a2 and x = ± a If we put x = 0 in Equation 3, we get y2 = - b2, which is impossible, so there -.is no y-intercept.The hyperbola consists of two parts, called its branches - The hyperbola have two asymptotes, which are the lines y = (b/a)x and y = - - (b/a)x shown in Figure 1.26. Both branches of the hyperbola approach the asymptotes; that is, they come arbitrarily close to the asymptotes. This.involves the idea of a limit, which is discussed in proceeding chapters Figure 1.26 2 9 29  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department By interchanging the roles of and we get an equation of the form 𝑦2 𝑥2 − =1 𝑎2 𝑏2.which also represents a hyperbola and is sketched in Figure 1.27 Figure 1.27.Example 24: Sketch the curve 9x2 - 4y2 = 36 :Solution: Dividing both sides by 36, we obtain 𝑦2 𝑥2 1= − 9 4 which is the standard form of the equation of a hyperbola. Since a2 = 4, the x- intercepts are ± 2. Since b2 = 9, we have b = 3 and the asymptotes are y = ± (3/2) x..The hyperbola is sketched in Figure 1.28 3 0 30  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.28 If b = a, a hyperbola has the equation x2 – y2 = a2 (or y2 – x2 = a2) and is called an equilateral hyperbola [see Figure 1.29(a)]. Its asymptotes are y = ± x, which are. perpendicular If an equilateral hyperbola is rotated by 45 , the asymptotes become the x- and y- 0 on of the hyperbola is xy = k, where k equati is aaxes, and it can be shown that the new constant [see Figure 1.29(b)] Figure 1.29 3 1 31  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Functions; Domain and Range 1.3 Functions arise whenever one quantity depends on another. A function can be.represented by an equation, a graph, a numerical table, or a verbal description We usually consider functions for which the sets D and E are sets of real numbers.. The set D is called the domain of the function ”.The number f(x) is the value of f at x and is read “f of x The range of f is the set of all possible values of f(x) as x varies throughout the. domain A symbol that represents an arbitrary number in the domain of a function f is called. an independent variable A symbol that represents a number in the range of f is called a dependent.variable Thus we can think of the domain as the set of all possible inputs and the range as the set of all possible outputs if we see the function as a kind of machine (Figure.1.30) Figure 1.30.Example 25: Verify the domains and associated ranges of the following functions y = x2 ) a ( -( The formula y = x2 gives a real y-value for any real number x, so the domain is The range of y = x2 is [0, ∞) because the square of any real number.)∞ ,∞.is nonnegative and every nonnegative number y is the square of its own square root 3 2 32  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department y = 1/x )b( The formula y = 1/x gives a real y-value for every x except x = 0. For consistency in the rules of arithmetic, we cannot divide any number by zero. The range of y = 1/x, the set of reciprocals of all nonzero real numbers, is the set of all nonzero real numbers, since y = 1/(1/y).That is, for y ≠ 0 the number x = 1/y is the input.assigned to the output value y y = √𝑥 ) c ( The formula y = √𝑥 gives a real y-value only if x ≥ )∞ ,0. The domain of y = √𝑥 is [0 The range of y = √𝑥 is [0, ∞) because every nonnegative number is some.number’s square root y = √4 − 𝑥 )d( The quantity 4 – x cannot be negative. That is, 4 – x ≥ 0, or x ≤ 0. The formula gives real y-values for all x ≤ 4. The domain is (-∞, 4] The range of function is [0, ∞), the set of all nonnegative.numbers y = √1 − 𝑥 2 )e( The domain is [-1, 1] The range is [0, 1] Graphs of Functions 1.3.1 The most common method for visualizing a function is its graph. If f is a function with domain D, then its graph is the set of ordered pairs In other words, the graph of f ).Notice that these are input-output pairs( consists of all points (x, y) in the coordinate plane such that y = f.(x) and x is in the domain of f 3 3 33  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department.Example 26: Graph the function y = x2 over the interval [-2, 2] :Solution.Make a table of xy-pairs that satisfy the equation y = x2.1 Plot the points (x, y) whose coordinates appear in the table (see Figure.2 1.31) Figure 1.31 3 4 34  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Draw a smooth curve (labeled with its equation) through the plotted points..3 (Figure 1.32) Figure 1.32 Piecewise-Defined Functions 1.3.2 Sometimes a function is described by using different formulas on different parts of. its domain. One example is the absolute value function 𝑥≥ 𝑥− 0 x|= | , 𝑥 { , 𝑥< ,0 side of the equation means that whose graph is given in Figure 1.33. The right-hand. the function equals x if x ≥ 0, and equals - x if x ˂ 0 3 5 35  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.33 Example 27: The function 𝑥− 𝑥 ,1 1:The values of ƒ are given by ,when x ˂ 0y = -x when 0 ≤ x ≤ 1 andy = x2 when x ˃ 1y = 1 The function, however, is just one function whose domain is the entire set of real numbers (see Figure 1.34) Figure 1.34 Example 28: A function is defined by 𝑥≥ 1− f(x) =𝑥{ 2 0 , ,𝑥 𝑥< Evaluate f(0), f(1)and f(2) and sketch the graph. ,0 : Solution Since 0 ≤ 1, we have f (0) = 1 – 0 = 1 Since 1 ≤ 1, we have f (1) = 1 – 1 = 0 Since 2 ˃ 1, we have f (2) = 22 = 4 3 6 36  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department See Figure 1.35 Figure 1.35 Increasing and Decreasing Functions 1.3.3 The graph shown in Figure 1.36 rises from A to B, falls from B to C, and rises again from C to D. The function f is said to be increasing on the interval [a, b],.decreasing on [b, c], and increasing again on [c, d] Figure 1.36 3 7 37  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Example 29: investigate the increasing and decreasing intervals of the functions y x2 = :Solution In the definition of an increasing function it is important to realize that the inequality f(x1) ˂ f(x2) must be satisfied for every pair of numbers x1 and x2 in I with. x2 ˂ x1 )∞ ,For interval [0 f(1) = 1 f(2) = 4 ,So that, according to the 1 definition the function is increasing on the interval [0 st )∞ ]For interval (- ∞, 0 f(-1) = 1 f(-2) = 4 The function is decreasing on the interval (- ∞, 0] See Figure 1.37 Figure 1.37 3 8 38  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Even Functions and Odd Functions: Symmetry 1.3.4.The graphs of even and odd functions have characteristics symmetry properties ,Generally The graph of an even function is symmetric about the y-axis. As for the function f(x) = x2 (see Figure 1.38) Since always f(-x) = f(x) Figure 1.38 The graph of an odd function is symmetric about the origin. For example the function y = x3 (Figure 1.39) Always f(-x) = - f(x) Figure 1.39 3 9 39  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Example 30: Determine whether each of the following functions is even, odd, or. neither even nor odd (c) h(x) = 2x – x (b) g(x) = 1 – x4f(x) = x5 + x )a( 2 :Solution )a( f(-x) = (-x)5 + (-x) = (- 1)5x5 + (-x) = - x 5 – x = - ( x 5 + x ) = - Figure 1.40 f ( x ) 4 Therefore 0 40 is an odd.function )b (  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Trigonometric Functions.5 Angles )a( Angles can be measured in degrees or in radians (abbreviated as rad). The angle given by a complete revolution contains 360o, which is the same as 2π rad. Therefore and rad =) (𝜋 180 0 ≈1 =( 1)o 180 57.3o 𝜋 0 rad ≈ 0.017o :Example 31.Find the radian measure of 600 )a(.Express 5π/4 rad in degrees )b( :Solution From above equations we see that to convert from degrees to radians we )a( rad = ) ( 60 = 60° multiply by π/180. Therefore 𝜋 𝜋 180 3 5𝜋 5𝜋 180 To convert from radians to degrees we multiply by 180/ π. Thus )b( 𝑟𝑎𝑑 = ( ) = 225° 𝜋 4 4 Table 1.2 shows the equivalence between degree and radian measures for some.basic angles 4 1 41  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.41 shows a sector of a circle with central angle θ and radius r subtending.an arc with length a Figure 4.41 Since the length of the arc is proportional to the size of the angle, and since the 𝑎𝜃 : entire circle has circumference 2πr and central angle 2π, we have = 2𝜋𝑟2𝜋 Solving this equation for θ and for a, we obtain.Remember that the above equations are valid only when is measured in radians :Example 32 ?If the radius of a circle is 5 cm, what angle is subtended by an arc of 6 cm )a( If a circle has radius 3 cm, what is the length of an arc subtended by a central )b( ?angle of 3π/8 rad :Solution Using Equation 𝜃 = and 𝑎 = 𝑟𝜃 with a = 6 and r = 5, we see that the )a( 𝑎 𝑟 θ = 6/5 = 1.2 rad angle is With r = 3 cm and θ = 3π/8 rad, the )b( :arc length is 4 2 42  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department 𝑎 = 𝑟𝜃 = 3 ( ) = 3𝜋 9𝜋 cm 8 8 standard position)b( The standard position of an angle occurs when we place its vertex at the origin of. a coordinate system and its initial side on the positive x-axis as in Figure 4.42 Figure 4.42: θ ≥ 0 A positive angle is obtained by rotating the initial side counterclockwise until it coincides with the terminal side. (as in Figure 4.42) A negative angle are obtained by clockwise rotation as in Figure 4.43 Figure 4.43: θ ˂ 0 Figure 4.44 shows several examples of angles in standard position. Notice that different angles can have the same terminal side. For instance, the angles 3π/4, -5 :π/4 and 11π/4 have the same initial and terminal sides because 4 3 43  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department − 2𝜋 = − 3𝜋 3𝜋 4 4 + 2𝜋 = 5𝜋.and 2π rad represents a complete revolution 11𝜋 4 4 Figure 4.44 The Six Basic Trigonometric Functions)c( For an acute angle θ the six trigonometric functions are defined as ratios of lengths. of sides of a right triangle as follows (see Figure 1.45) Figure 1.45 4 4 44  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department We extend this definition to obtuse and negative angles by first placing the angle in.standard position in a circle of radius r We then define the trigonometric functions in terms of the coordinates of the point.P(x, y) where the angle’s terminal ray intersects the circle (Figure 1.46) Figure 1.46 These extended definitions agree with the right-triangle definitions when the angle. is acute ,Notice also that whenever the quotients are defined As you can see, tan θ and sec θ are not defined if x = cos θ = 0. …, This means they are not defined if θ is ±π /2, ±3π/2 Similarly, cot θ and csc θ are not defined for values of θ for which y = 0, namely θ …,π, ±2π± ,0 = The exact trigonometric ratios for certain angles can be read from the triangles in ,Figure 1.47. For instance 4 5 45  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.47 The signs of the trigonometric functions for angles in each of the four quadrants can be remembered by means of the rule “All Students Take Calculus” or “CAST”.rule shown in Figure 1.48 Figure 1.48.Example 33: Find the exact trigonometric ratios for θ = 2π/3 :Solution: From the triangle in Figure 1.49 we see that tan = −√3 cos = − , sin = 2𝜋 2𝜋 1 2𝜋 √3 2 3 2 3 , 3 4 6 46  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.49 Using a similar method we determined the values of sin θ, cos θ, and tan θ shown in Table 1.3 Table 1.3 Example 34: If cos θ = 2/5 and 0 ˂ θ ˂ π/2, find the other five trigonometric.functions of θ :Solution Since cos θ =2/5, we can label the hypotenuse as having length 5 and the adjacent side as having length 2 in Figure 1.50. If the opposite side has length x, then the.Pythagorean Theorem gives x2 + 4 = 25 and so x2 = 21, x = √21 4 7 47  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Figure 1.50 :We can now use the diagram to write the other five trigonometric functions tan 𝜃 = √21 sin 𝜃 = √21 5 2 csc 𝜃 5 = cot 𝜃sec 𝜃 = 2 5 2 = 21√ 21√.Example 35: Use a calculator to approximate the value of x in Figure 1.51 Figure 1.51 :Solution: From the diagram we see that tan 40° = 16 1 ,Therefore 𝑥 6 ≈ tan = 19.07 40° 4 8 48  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Trigonometric Identities)d( A trigonometric identity is a relationship among the trigonometric functions. : The most elementary are the following For the next identity we refer back to Figure 1.46. The distance formula (or, equivalently, the Pythagorean Theorem) tells us that x2 + y2 = r2. Therefore 𝑥2 + + 2 = 𝑠i𝑛 𝜃 + 𝑥2𝑐𝑜𝑠 𝑦 𝜃 𝑟2 2 2 2 𝑟 𝑟2 𝑦2 == = 𝑟2 𝑟2 1 :We have therefore proved one of the most useful of all trigonometric identities This equation, true for all values of θ, is the most frequently used identity in. trigonometry :Dividing this identity in turn by cos2 θ and sin2 θ gives The identity 4 9 49  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department.Show that sin is an odd function and cos is an even function :Since the angles θ and θ + 2π have the same terminal side, we have The remaining trigonometric identities are all consequences of two basic identities : called the addition formulas By substituting – y for in above equations and using equations [sin (-θ) = -sin (θ) :and cos (-θ) = cos (θ)] we obtain the following subtraction formulas Then, by dividing the formulas in addition formulas or subtraction formulas, we : obtain the corresponding formulas for tan (x ± y) 5 0 50  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department :If we put y = x in the addition formulas, we get the double-angle formulas Then, by using the identity sin2x + cos2x = 1, we obtain the following alternate : forms of the double-angle formulas for cos 2x If we now solve these equations for cos2x and sin2x, we get the following half- :angle formulas, which are useful in integral calculus Finally, we state the product formulas, which can be deduced from addition and : subtraction formulas.Example 36 Find all values of x in the interval [0, 2π] such that sin x = sin 2x Solution: Using the double-angle formula, we rewrite the given equation as sin x(1 – 2cos x) =or sin x = 2 sin x cos x 0 5 1 51  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department :Therefore, there are two possibilities sin x = 0 1 – 2cos x = 0or x = 0, π, 2π ½ = cos x x = π/3, 5π/3.The given equation has five solutions:0, π/3, π, 5π/3, and 2π Periodicity and Graphs of the Trigonometric Functions)e( The graph of the trigonometric function is obtained by plotting points for one. period and then using the periodic nature of the function to complete the graph We describe this repeating behavior by saying that the six basic trigonometric :functions are periodic Periods of Trigonometric Functions tan (x + π) = tan x Period π: cot (x + π) = cot x sin (x + 2π) = sin xPeriod 2π: cos (x + 2π)= cos x sec (x + 2π)= sec x csc (x + 2π) = csc x.Example 37: plot the trigonometric function sin x :Solution The function will be: y = f (x) = sin (x) Domain: - ∞ ˂ x ˂ ∞ Range: -1 ≤ y ≤1 5 2 52  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department Make a table for xy values x y = sin (x) 0 0 π/2 1 π 0 3 π/2 -1 2π 0 - π/2 -1 -π 0 The plot of sin x was shown in Figure 1.52. The function y = sin x is an odd function ( f (-x) = - f (x)) Figure 1.52 Figure 1.53 shows the graphs of the six basic trigonometric functions using radian. measure. The shading for each trigonometric function indicates its periodicity 5 3 53 Calculus Lectures Tikrit University 1 st class College of Engineering Ch.1 Preliminaries Civil Engineering Department 5 4 54  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department CHAPTER 2 Limits and Continuity Limits 2.1 If the values of f (x) tend to get closer and closer to the number L as x gets closer. and closer to the number a (from either side of a) but x ≠ a.Example 1: Find the limit of the function f (x) = x2 – x +2 as x approaches 2 :Solution Let’s investigate the behavior of the function f (x) = x – x +2 for values of x near 2 The following table gives values of f (x) for values of x close to 2, but not equal.2. to 2 1 55  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department Figure 1 From the table and the graph of f (a parabola) shown in Figure 1 we see that when x is close to 2 (on either side of 2), f (x) is close to 4. In fact, it appears that we can make the values of f (x) as close as we like to 4 by taking x suffi ciently close to 2. We express this by saying “the limit of the function f (x) = x2 – x +2 as x ”.approaches 2 is equal to 4 :The notation for this is lim(𝑥2 - 𝑥 + 2) = 4 𝑥→ 2 Example 2: Guess the value of lim𝑥→1 𝑥−1 𝑥 2 −1 :Solution Notice that the function f (x) = (x – 1) / (x – 1) is not defined when x =1, but that 2 doesn’t matter because the definition of limx→a f (x) says that we consider values of.x that are close to a but not equal to a The tables below give values of f (x) (correct to six decimal places) for values of x.that approach 1 (but are not equal to 1) 2 56  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department On the basis of the values in the tables, we make the guess that lim𝑥→1 𝑥−1 𝑥 2 −1 = The value of lim𝑥𝑥→1 𝑥−1 0.5 2 −1 can be solved to give same results 𝑥 −( :by 𝑥−1 lim = )1 𝑥→1 (𝑥 − 1)(𝑥 +𝑥→1 𝑥 2 − 1 lim 1 li 1) 𝑥→1 𝑥 + m1 = 0.51 lim 𝑥→1 (1) + 1.Example 2 is illustrated by the graph of in Figure 2 Figure 2 3 57  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department Example: 3 ,If ƒ is the identity function f (x )= x, then for any value of x0 (Figure 3a) )a( lim ƒ(𝑥) = lim 𝑥 = 𝑥0 𝑥→𝑥 0 𝑥→𝑥 0 If ƒ is the constant function f (x) = k (function with the constant value)b( ,k), then for any value of x0 (Figure 3b) lim ƒ(𝑥) = lim 𝑘 = 𝑘 𝑥→𝑥 0 𝑥→𝑥 0 Figure 3 Example 4: Find lim𝑥→0 if it 1 𝑥2.exists Solution: As x becomes close to 0, x2 also becomes close to 0, and 1/x2 becomes.very large. (See the table) Figure 4 4 58  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department In fact, it appears from the graph of the function f(x) =1/x2 shown in Figure 4 that the values of f (x) can be made arbitrarily large by taking x close enough to 0. Thus.the values of f (x) do not approach a number, so limx→0 (1/x2) does not exist :To indicate the kind of behavior exhibited in Example 4, we use the notation 1 =𝑥→0lim 𝑥2 ∞ Example 5: Investigate lim𝑥 →0 sin 𝜋 :Solution Again the function f (x) = sin (π/x) is undefined at 0. The graph of function was. shown in Figure 5 Figure 5 The graph indicate that the values of sin (π/x) oscillate between 1 and -1 infinitely. often as x approaches 0 ,Since the values of do not approach a fixed number as approaches 0 do not existlim𝑥→0 sin 𝜋 𝑥 5 59  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department The Limit Laws 2.1.1 To calculate limits of functions that are arithmetic combinations of functions :having known limits, we can use several easy rules Example 6: Use the observations limx→c k = k and limx→c x = c (Example 3) and :the properties of limits to find the following limits (b) lim𝑥→𝑐lim𝑥→𝑐 ( 𝑥 + 4𝑥 − 3) )a( 3 2 𝑥 4 +𝑥 2 − )c( 𝑥 →− 4𝑥2 −√ 𝑥 2 +5 1 :Solution lim 2 3 6 60  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department :Example 7: The following calculation illustrates Theorems 2 and 3 0 (−1)3 + 4(−1)2 − 3𝑥 3 + 4𝑥 2 − 3 li = 0 == 2 (−1) + 5 +2 5𝑥 6 m1 𝑥 →− 7 61  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department Example 8: Evaluate 𝑥 → 𝑥 +100−1√0 2 lim 0 𝑥2 Solution: We can create a common factor by multiplying both numerator and denominator by the conjugate radical expression √𝑥 2 + 100 + 10 (obtained by changing the sign after the square root). The preliminary algebra rationalizes the : numerator :Indeterminate Forms.2 :There are seven indeterminate forms ∞ and 1 ,0∞ ,00 ,∞-∞ ,∞.0 ,∞/∞ ,0/0 Sandwich Theorem.3 The following theorem enables us to calculate a variety of limits. It is called the Sandwich Theorem because it refers to a function ƒ whose values are sandwiched between the values of two other functions g and h that have the same limit L at a point c. See Figure 6 8 62  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department Figure 6 The Sandwich Theorem is also called the Squeeze Theorem or the Pinching.Theorem Example 9: Given that ≤ u (x) ≤ 1 + −1 𝑥2 𝑥2 for all x ≠ 0 2 4.find limx→0 u(x), no matter how complicated u is Solution: Since ,and limx→0 (1 + (x /2)) = 1limx→0 (1 – (x /4)) = 1 2 2.the Sandwich Theorem implies that limx→0 u(x) = 1 (Figure 7) 9 63  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department Figure 7 Example 10: Show that lim𝑥→0 𝑥 2 sin = 1 𝑥 Solution: First note that we cannot use.0 1 0 64  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department Figure 8 One-Sided Limits 2.1.4 In this section we extend the limit concept to one-sided limits, which are limits as x approaches the number c from the left-hand side (where x ˂ c) or the right-hand.side (x ˃ c) only :In another word Right-hand limit is the limit of f(x) as x approaches c from the right, or  lim𝑥→𝑐+ ƒ(𝑥) Left-hand limit is the limit of f(x) as x approaches c from the left, or  lim𝑥→𝑐− ƒ(𝑥) lim𝑥→𝑐 ƒ(𝑥) = L if and only if lim𝑥→𝑐+ ƒ ( 𝑥 ) = 𝐿 and lim𝑥→𝑐− ƒ(𝑥) =𝐿  3 𝑥< f (x)= {,𝑥 𝑥 − 2 Example 11: Let 2 + 𝑥> ,1 2 1 1 65  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department Find lim𝑥→2+ ƒ(𝑥) and lim𝑥→2− )a( ƒ(𝑥) Does lim𝑥 →2 ƒ (𝑥) exist ? )b( ? why lim𝑥→2+ ƒ(𝑥) = lim𝑥→2+ ƒ( 2 :+Solution 1) = 2/2 + )a( 1=2 lim𝑥 →2− ƒ (𝑥) = lim𝑥 →2− (3 − 𝑥) = 3 – 2 = 1 lim𝑥 →2+ ƒ (𝑥) ≠ lim𝑥 →2− )b( ƒ (𝑥) lim 𝑥 G does not exist ƒ(𝑥) ,Example 12: let f (x)= {1 − 𝑥𝑥→2 𝑥 = 1,2 1 2 Find lim𝑥 →1+ ƒ (𝑥) and lim𝑥 →1− ƒ (𝑥) )a( ? Does lim𝑥→1 ƒ(𝑥) exist ? why )b( Graph f (x) )c( :Solution lim𝑥 →1+ ƒ (𝑥) = lim𝑥 →1+ 1 − 𝑥 2 = 1 – (1)2 = 0 )a( lim𝑥→1− ƒ(𝑥) = lim𝑥→1− 1 − 𝑥 2 = 1 – (1)2 = 0 lim𝑥→1 ƒ(𝑥) existlim𝑥→1+ ƒ(𝑥) = lim𝑥→1− ƒ(𝑥), )b( Limits Involving (sin θ/θ) 2.1.5.A central fact about (sin θ/θ) is that in radian measure its limit as θ → 0 is 1 Proof The plan is to show that the right-hand and left-hand limits are both 1. Then. we will know that the two-sided limit is 1 as well 1 2 66  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department To show that the right-hand limit is 1, we begin with positive values of less than.π/2 (Figure 9) Figure 9 1 3 67  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department :Since limθ→0+ cos θ = 1, the Sandwich theorem gives Figure 10 Example 13: show that (b) lim𝑥 →0 = and sin 2𝑥 2 limℎ →0 = )a ( co s ℎ −1 ℎ 5 5𝑥 0 :Solution 1 4 68  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department Example 14: Find lim𝑡 →0 tan 𝑡 sec 3 2𝑡 𝑡 Limits at 2.1.6 Infinity lim𝑥𝑥→−∞ = 0lim𝑥→∞ = 0, :General rules 1 1 𝑥 lim𝑥→∞ 𝑘 lim𝑥→−∞ 𝑘 ,= 𝑘 =𝑘 lim𝑥→∞ 𝜃 = 0, to prove 𝑠i𝑛𝜃 :it 𝑠i𝑛𝜃 1− sin θ ≤ 1 ≤ 1- ≤ ≤ 1 lim𝜃→∞ = 0 and𝜃 lim = 0 , then from Sandwich [÷ θ] −1 1 𝜃𝜃→∞ 𝜃 :theorem 𝜃 𝜃 𝑠i𝑛𝜃 lim → 𝜃 = ∞ 0 1 5 69  Calculus Lectures Tikrit University 1 st class College of Engineering Ch.2 Limits and Continuity Civil Engineering Department Limits at Infinity of Rational Functions :There are two methods Divide both the numerator and the denominator by the highest power of x in.1. denominator.Suppose that x = 1/h and find limit as h approaches zero.2 ƒ(𝑥) g (𝑥 Note: for rational function ) If degree of f(x) less than degree of g(x), then lim𝑥→±∞ =.1 ƒ(𝑥) g (𝑥 ) 0 If degree of f(x) equals degree of g(x), then lim𝑥→±∞ ƒ(𝑥) g (𝑥 is.2 ) If degree of f(x) greater than degree of g(x), then lim𝑥→±∞ ƒ(𝑥).finite g (𝑥 is.3 ).infinite 𝑥 3 −4𝑥 2 + →𝑥 2𝑥 2 − 7 lim Example 15: find ∞ 3 = lim 𝑥 3 4𝑥 2 7 𝑥 2− + 𝑥2 𝑥−4+ 7 ∞ = 2 = 𝑥2−0 𝑥2 ∞ ∞−4+0 →𝑥 →𝑥 2 lim Method 1: 2𝑥 2 3 3 ∞ ∞ −2 = − 2 𝑥2𝑥2 𝑥 Method 2: let x = 1/h, h → 0 = lim +7 − 2 1 4 1 ℎ3 li ℎ→ 3 1 1−4ℎ+7ℎ 3 ℎ2 4( ℎ ) +7 − )ℎ ( m 0 2 −3) 1 ( 2 ℎ→0 2 −3 ℎ→0 ℎ 32 2−3ℎ ℎ2 = ℎ ℎ2 lim li 1−4ℎ+7ℎ3 1−4(0)+7(0) 1 m0 0(2−3(0))ℎ→0 ℎ(2−3ℎ2 ) = = = 𝑥 3/2 +5 lim Example 16: Find ∞ 𝑥→∞ √𝑥 3 +4 𝑥 3/2 = lim𝑥 = 5 5 + +1 1 lim𝑥→∞ 𝑥 3/2 𝑥 3/2 ∞→ 𝑥 3/2 = 1= 1√ 1+0√ 𝑥3 4 + 3√ 4 + 1𝑥√ 3 1+0 𝑥 𝑥3 1 6 70  Calculus Lectures Tikrit University 1 st class