Limits of Functions PDF

cHAP 5: LIMITS OF FUNCTIONS Perhaps you have heard about speed limit. What does the word “limit” means to you?? Or maybe you have been in a parking-lot situation in which you must “inch up” to the car in front, but yet you do not want to bump or touch it. This notation of getting closer and closer...

cHAP 5: LIMITS OF FUNCTIONS Perhaps you have heard about speed limit. What does the word “limit” means to you?? Or maybe you have been in a parking-lot situation in which you must “inch up” to the car in front, but yet you do not want to bump or touch it. This notation of getting closer and closer to something, but yet not touching it, is very important in mathematics and is involved in the concept of limits. Basically, we will let a variable “inch up” to a particular value and examine the effect it has on the values of a function. PARKING SPACE 5.1 DEFINITION OF LIMITS  Limits is based on the idea “getting closer to something” or “approaching something but yet not touching it”.  The limit of f(x) as x approaches c is the number L, written lim f ( x)  L x c provided that f(x) is arbitrarily close to L for all x sufficiently close to, but not equal to, c.  We emphasize that, when finding a limit, we are concerned not with what happens to f(x) when x equals c, but only with what happens to f(x) when x is close to c. 1 of 22  cHAP 5: LIMITS OF FUNCTIONS  Moreover, a limit must be independent of the way in which x approaches c. That is, the limit must be the same whether x approaches c from the left or from the right(for x < c or x > c, respectively).  For example: Let say; ƒ(x) = 2x + 1, the limit for this function as x is approaching the value of 1 is denoted by: lim f x   lim 2 x  1 x 1 x 1  In this case, x can approach 1 from the left and also from the right: i) Left : denoted by x 2 , means x can take values that are less than and close to 2 (1.9, 1.99, 1.999, 1.9999….etc) ii) Right: denoted by x 2 , means that x can take values that are greater and closer to 2 (2.1, 2.01, 2.001, 2.0001…)  There are three ways for us to obtain the limits of a function: From the table From the graph Using Algebra 5.1.1 ESTIMATING LIMIT OF A FUNCTION: FROM THE TABLE Suppose that we are given a function, f ( x)  3 x 2  1 and we would like to examine what is the value (limit) of this function as the input (x) approaches 4. In other words, we try to solve this question;  lim f x   lim 3x 2  1 x 4 x 4  As we know, x can approach 4 from the left and also from the right; 2 of 22  cHAP 5: LIMITS OF FUNCTIONS Step 1: x approaches 4 from left: x 4 : This means the values of x must be less than but close to 4: x  3.9 , x  3.99 , x  3.999 , x  3.9999 Closest to 4 Closer to 4 Closer to 4 Less than 4 Substitute the selected values of x to examine the effect on f x : f  x   3x 2  1 When x  3.9 f 3.9  33.9  1  44.63 2 f  x   3x 2  1 When x  3.99 f 3.99  33.99  1  46.7603 2 x  3.999 and Carry on with values of x that are closer to 4 like x  3.9999. What can you conclude??  as x approach 4 (from left) f(x) will approach …….. You can also examine the limit for the function using a table: Table: x 4 x 3.9 3.99 3.999 3.9999 f  x  =3x2 -1 As x 4 , we observe that f ( x) ??? 3 of 22  cHAP 5: LIMITS OF FUNCTIONS From the table, we can see that, as x approaches 4 from the left, f x is getting closer to (approaching) the value of 47 or  lim f x   lim 3x 2  1  47 x 4 x 4  Step 2: Similarly, now let us examine if x approach 4 from the right: x 4 where the values of x must be greater than 4. Select a few values of x that are greater than 4 (but remember it must be close to 4) like x  4.1, x  4.01, x  4.001 and x  4.0001. Calculate the value of f(x) for each x. Then, examine the effects on f(x) from the table: Table: x 4 x 4.1 4.01 4.001 4.0001 f  x  =3x2 -1 As x 4 , we observe that f ( x) ??? From the table, we can see that, as x approaches 4 from the right, f x is getting closer to (approaching) the value of 47 or  lim f x   lim 3x 2  1  47 x 4 x 4  Step 3: EQUAL Since lim f x   lim f x   47 x 4  x 4  Therefore we can conclude x 4  lim f x   lim 3x 2  1  47 x 4  4 of 22  cHAP 5: LIMITS OF FUNCTIONS DEFINITION 5.1.1(a): The limit of a function (as x approaches a) exists, if and only if the limit of the function from both sides (left & right) are equal. If NOTES: lim f x   lim f x  , x a  x a then lim f x  DOES NOT EXIST xa a  any value Example 1: x3  1 Given f x  , find the limit for f  x  as x approaches 1. x 1 Solution: Step 1: Build the table for x approaching 1 from the left : x  1. Calculate the value of f(x) for each x. Table : x 1 x 0.9 0.99 0.999 0.9999 f x As x 1 , we observe that f ( x) ??? 5 of 22  cHAP 5: LIMITS OF FUNCTIONS As x is approaching 1 from the left, we can see that f x is approaching 3 or:  x3  1  lim f x   lim    3 x1 x1  x 1  Step 2: Build the table for x approaching 1 from the right : x  1. Calculate the value of f(x) for each x. Table : x 1 x 1.1 1.01 1.001 1.0001 f x As x 1 , we observe that f ( x) ??? As x is approaching 1 from the right, we can see that f x is approaching 3 or:  x3  1  lim f x   lim    3 x1 x1  x  1  Step 3: EQUALS As lim f x   lim f x   3 x1 x1  x3 1 Therefore, lim f x   lim    3 x 1 x 1  x  1  6 of 22  cHAP 5: LIMITS OF FUNCTIONS Example 2: 1 lim Determine whether x 3  x  3 exist or not? Solution: Step 1: Build the table for x approaching 3 from the left : x  3. Calculate the value of f(x) for each x. Table : x 3 x 2.9 2.99 2.999 2.9999 f x As x 3 , we observe that f ( x) ??? As x is approaching 3 from the left, we can see that f x is getting smaller and smaller, therefore lim f x   lim 1   x3 x3 x  3 (i.e. As x 3 , the function f (x)  ) 7 of 22  cHAP 5: LIMITS OF FUNCTIONS Step 2: Build the table for x approaching 3 from the right : x  3. Calculate the value of f(x) for each x. Table : x 3 x 3.1 3.01 3.001 3.0001 f x As x 3 , we observe that f ( x) ??? As x is approaching 3 from the right, we can see that f x is getting larger and larger, therefore: lim f x   lim 1  x3 x3 x  3 (i.e. As x 3 , the function f (x)  ) NOT EQUAL Step 3: As lim f x   lim f x  x 3 x 3 lim f x   lim 1  DOES NOT EXIST Therefore x 3 x 3  x  3 8 of 22  cHAP 5: LIMITS OF FUNCTIONS 5.2 ESTIMATING LIMIT OF A FUNCTION: FROM THE GRAPH Another way to obtain the limit of a function is from the graph. Example 3: Based on the graph for the function f  x  , we can examine the limit for this function as x approach 1: y 6 5 4 3 2 1 -4 -3 -2 -1 0 1 2 3 x x 1 1  x   a) Find i) lim f x  ii) lim f x  x 1 x 1 iii) lim f x  x 1 9 of 22  cHAP 5: LIMITS OF FUNCTIONS Note: To understand how to find a)(i) lim f x  from the graph of f(x), you may x 1 also build the table for x approaching 1 from the left : x  1. Then, you can read the (approximate) value of f(x) from the graph for each x. Table : x 1 x 0.9 0.99 0.999 0.9999 f x As x 1 , we observe that f ( x) ??? As x is approaching 1 from the left, we can see that f x is approaching 2, that is, lim f x   2 x1 Step 2 in the table method can also be used where you can read the values of f(x) from the graph for each x to find a)(ii) lim f x  x 1 Table : x 1 x 1.1 1.01 1.001 1.0001 f x As x 1 , we observe that f ( x) ??? As x is approaching 1 from the right, we can see that f x is approaching 4 that is, lim f x   4 x 1 Step 3 in the table method is used to find a)(iii) lim f x  x 1 NOT EQUAL Since lim f x   lim f x  x 1 x 1 Therefore, lim f x  does not exist x 1 10 of 22  cHAP 5: LIMITS OF FUNCTIONS Example 4: y 6 5 4 3 2 1 -4 -3 -2 -1 0 1 2 3 x x 2 2  x   b) Find i) lim f x  ii) lim f x  x 2  x 2  iii) lim f x  x 2 11 of 22  cHAP 5: LIMITS OF FUNCTIONS Example 5: y 6 5 4 3 2 1 -4 -3 -2 -1 0 1 2 3 x x 1  1  x   c) Find i) lim f x  ii) lim f x    x 1   x 1 iii) lim f x  iv) f  1 x  1 12 of 22  cHAP 5: LIMITS OF FUNCTIONS 5.3 ESTIMATING LIMIT OF A FUNCTION: USING ALGEBRA PROPERTIES OF LIMITS: PROPERTY 1 lim c  c i) x a where c is a constant Example 6: a) lim 7  x2 b) lim e  x 5 PROPERTY 2 lim x n  a n ii) xa Example 7: a) x 3 lim x 2  b) lim x1 / 3  x 27 13 of 22  cHAP 5: LIMITS OF FUNCTIONS Given lim f x   L and lim g x   M x a x a where L and M are real numbers, then: PROPERTY 3 lim f x   g x   lim f x   lim g x  iii) x a x a x a = L ± M PROPERTY 4 lim f x .g x   lim f x . lim g x  iv) x a x a x a =LxM PROPERTY 5 limc. f x   c. lim f x  v) x a x a =cL PROPERTY 6 lim n f x   n lim f x  vi) x a x a n = L 14 of 22  cHAP 5: LIMITS OF FUNCTIONS Example 8: a) Given f ( x)  4 x  3. Using the properties of limits, then lim f ( x)  lim (4 x  3) x 2 x 2  lim 4 x  lim 3 ( property3) x 2 x2  4 lim x  lim 3 ( property5) x 2 x2  4(2)  3 ( property 2 & property1)  11 b) Given g ( x)  x 2  1. Using the properties of limits, then lim g ( x)  lim ( x 2  1) x 3 x 3  lim ( x 2  1) ( property6) x 3  lim x  lim 1 x 3 2 x 3 ( property3)  3)    1  ( property2 & property1) 2  10 Notice that lim (4 x  3)  4(2)  3 x 2 and lim ( x 2  1)  (3)2  1 x 3 Therefore, direct substitution of x=a into f(x) may yield the limit value. 15 of 22  cHAP 5: LIMITS OF FUNCTIONS Example 9: Given f ( x)  2 x  3 and g ( x)  x 2  1. First, find the following limits using the properties of limits. Then, use direct substitution. a) lim f ( x)  x 2 b) lim g ( x)  x 2 c)  lim (2 x  3)  ( x 2  1)  x 2  d) lim (2 x  3)( x 2  1)  x 2 e) lim e( x 2  1)  x 2 f) lim (2 x  3)  x 2 16 of 22  cHAP 5: LIMITS OF FUNCTIONS PROPERTY 7 LIMIT PROPERTY FOR RATIONAL FUNCTIONS: If: lim f x   L and lim g x   M x a x a where L and M are real numbers, M ≠ 0, f x  h(x)  Then the limit for the rational function : g x  , is obtained by f x  L lim h( x)  lim  x a x a g x  M However, there are 4 possibilities for the values of L and M that will define the limits for this function: POSSIBILITY 1 L ≠ 0 and M ≠ 0 Therefore; f x  L lim  0 x a g x  M Example 10: Given: f ( x)  5 x  4 and g ( x)  3 x  7. Find a) lim (5 x  4)  x 3 b) lim (3x  7)  x 3 (5 x  4) c) lim  x 3 (3 x  7) 17 of 22  cHAP 5: LIMITS OF FUNCTIONS Example 11: Find: x2 a) lim x3 x 8 x2  5 b) lim  x2 1  x POSSIBILITY 2 L = 0 and M ≠ 0 Therefore; f x  0 lim  0 xa g x  M Example 12: x4 a) lim = x 4 x2 2x  4 b) lim  x2 3 POSSIBILITY 3 L ≠ 0 and M = 0 Therefore; f x  L lim  g x  0 DOES NOT x a EXIST 18 of 22  cHAP 5: LIMITS OF FUNCTIONS Example 13: x a) lim = x3 x 3 4x b) lim  x3 2 x  6 POSSIBILITY 4 L = 0 and M = 0 f x  0 0 lim  Indeterminate form: xa g x  0 0  When direct substitution of x with a yields an indeterminate form, check whether the function is in the simplest form or not. Simplify the function first using relevant methods such as factoring. Then, use suitable limit properties or simply use direct substitution again. Example 14: x2 lim a) x2  x2  4 3x 2  7 x  2 b) lim  x2 x4 19 of 22  cHAP 5: LIMITS OF FUNCTIONS 5.4 LIMITS OF FUNCTIONS AT INFINITY (∞) i) Limit at infinity for a constant function: PROPERTY 8 lim a  a where a is a constant. x  Limit of a constant function will always remain as the constant, regardless of the value that x is approaching to. Example 15: Find: a) lim   x  b) lim 2.718  x  ii) Limit at infinity for polynomial function: PROPERTY 9 lim ax n  a. n where a is constant. x To find limit at infinity for polynomial function, simply use direct substitution of x with . General guidelines of calculations with ∞ i) a+∞=∞ vi) a(-∞) = - ∞ ii) ∞+a=∞ vii) ∞/a = ∞ where a≠0 viii) ∞n = ∞; for n>0 iii) a - ∞ = -∞ iv) ∞-a=∞ ix) n  =∞ v) a.∞=∞ x) Indeterminate forms: 0  0  20 of 22  cHAP 5: LIMITS OF FUNCTIONS Example 16: Find; a) lim (3x 2  1)  x  b) lim (2 x 5  x 4  1000)  x 1 1 iii) Limit at infinity for functions: f ( x)  and f ( x)  p for p  0 : x x PROPERTY 10 1 1 lim 0 and lim 0 x  x x  x PROPERTY 11 1 lim  0 for p  0 x  x p Example 17: 1 1 1 a) lim   lim  property5 x  2 x 4 2  x x 4   1 0 property11 2 5 b) lim  x 2 x 1 c) lim  x  x 3 21 of 22  cHAP 5: LIMITS OF FUNCTIONS g ( x)  numerator iv) Limit at infinity for rational function: f ( x)  h( x )  deno min ator PROPERTY 12 If f(x) is a rational function and anxn and bmxm are the terms in the numerator and denominator, respectively, with the greatest powers of x, then an x n lim f ( x)  xlim  b x m and x  m an x n lim f ( x)  xlim  b x m x  m Steps to obtain limit at infinity for rational function: i. Exclude all terms in the numerator except the one term with the greatest power. Do the same for the denominator. ii. Simplify the new function. (To avoid indeterminate forms) iii. Substitute x with ∞ (or -∞) accordingly. iv. Calculate directly the limit value for the function or use suitable properties to obtain the value. Indeterminate forms: 0  Example 18: 0  2x4  x2  3 2x4 a) lim 3  lim 3  x x  x  2 x  x 2 x3  x 2  3 b) lim 3  x x  x  2 x 1 c) lim  x 2 x 2  5 x 22 of 22