MATH 102 Calculus 1 - Evaluating Limits PDF

Summary

These are lecture notes for a calculus course (MATH 102) at the Polytechnic University of the Philippines. The lecture notes cover evaluating limits, including limit theorems and examples. Topics covered include constant limits and the limit of identity functions.

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MATH 102 CALCULUS 1 Department of Mathematics College of Science Polytechnic University of the Philippines Sta. Mesa, Manila Evaluating Limits OUTLINE 1. Limit Theorems 2. Evaluating Limits 2 of 22 Evaluating Limits Limit The...

MATH 102 CALCULUS 1 Department of Mathematics College of Science Polytechnic University of the Philippines Sta. Mesa, Manila Evaluating Limits OUTLINE 1. Limit Theorems 2. Evaluating Limits 2 of 22 Evaluating Limits Limit Theorem 3 of 22 Evaluating Limits Limit Theorem 1. Uniqueness of Limits Let f be a function defined on some open interval I containing a, except possibly at a. If lim f (x) exists, then it is unique; that is, if lim f (x) = L1 x!a x!a and lim f (x) = L2 then L1 = L2. x!a 3 of 22 Evaluating Limits 2. Limit of a Constant If a, c 2 R , then lim c = c. x!a 4 of 22 Evaluating Limits 2. Limit of a Constant If a, c 2 R , then lim c = c. x!a 8 lim 5 = 5 7 X+ - 1 6 5 4 f lim 10 = 18 3 X+ 9 2 1 2024 lim +2024) = - 5 4 3 2 1 1 2 3 4 5 6 7 8 1 *- 0 2 3 4 5 4 of 22 Evaluating Limits 3. If f is the identity function, that is, f (x) = x, then lim x = a. x!a 5 of 22 Evaluating Limits 3. If f is the identity function, that is, f (x) = x, then lim x = a. x!a 8 f 7 6 lim x = 4 5 X+ 4 4 3 limX = -I 2 1 X+ - T 5 4 3 2 1 1 2 3 4 5 6 7 8 1 2 lim r = 0 3 r- 0 4 5 5 of 22 Evaluating Limits Let f and g be functions defined on some open interval I containing a, except possibly at a. 4. If lim f (x) and lim g (x) exist then x!a x!a 6 of 22 Evaluating Limits Let f and g be functions defined on some open interval I containing a, except possibly at a. 4. If lim f (x) and lim g (x) exist then x!a x!a a. (Limit of a Constant Multiple) lim c · f (x) = c · lim f (x), where c 2 R. x!a x!a lim 3X = 3 lim x X+ 4 * -> 4 = 3(4) = 12 6 of 22 Evaluating Limits Let f and g be functions defined on some open interval I containing a, except possibly at a. 4. If lim f (x) and lim g (x) exist then x!a x!a a. (Limit of a Constant Multiple) lim c · f (x) = c · lim f (x), where c 2 R. x!a x!a b. (Limit of Sum and Di↵erence) lim [f (x) ± g (x)] = lim f (x) ± lim g (x). x!a x!a x!a lim2 + lim lim (2 + x) = *+ I X X- - 1 X+ - 1 I 2 + (1) I I 6 of 22 Evaluating Limits Let f and g be functions defined on some open interval I containing a, except possibly at a. 4. If lim f (x) and lim g (x) exist then x!a x!a a. (Limit of a Constant Multiple) lim c · f (x) = c · lim f (x), where c 2 R. x!a x!a b. (Limit of Sum and Di↵erence) lim [f (x) ± g (x)] = lim f (x) ± lim g (x). x!a x!a x!a c. (Limit of Product) lim [f (x) · g (x)] = lim f (x) · lim g (x). x!a x!a x!a (mx +my) (3mx him (2(0) + 4) -1 =m(2x 4) m(3x = = (2x + 4)(3x 1) + x lim - - = X+ 0 x 6 of 22 Evaluating Limits f (x) lim f (x) d. (Limit of Quotient) lim = x!a , lim g (x) 6= 0. x!a g (x) lim g (x) x!a x!a im 7 of 22 Evaluating Limits f (x) lim f (x) d. (Limit of Quotient) lim = x!a , lim g (x) 6= 0. x!a g (x) lim g (x) x!a x!a set of natural numbers h in I e. (Limit of Power) lim [f (x)]n = lim f (x) , where n 2 N. x!a x!a 7 of 22 Evaluating Limits f (x) lim f (x) d. (Limit of Quotient) lim = x!a , lim g (x) 6= 0. x!a g (x) lim g (x) x!a x!a h in e. (Limit of Power) lim [f (x)]n = lim f (x) , where n 2 N. x!a x!a p q f. (Limit of Root) lim n f (x) = n lim f (x), where n 2 N, n > 1 and provided x!a x!a that lim f (x) > 0 when n is even. x!a 7 of 22 Evaluating Limits Assume the following: lim f (x) = 2, lim g (x) = 1, lim h(x) = 1 x!c x!c x!c Compute the following limits: lim g(x) + lim x 1. lim [5f (x) g (x) + 2h(x)] a g (x) + f (x) X+ C x!c 2. lim - x!c h(x) lim h(X) lim 2) -limg(x) + lim 5f(x) X+ C X- 2 X+ C -2) 1 felim - 5limf(x)-img(x) = X+ C = = 5(2) - (1) + 2(1) I = = - 13 8 of 22 Evaluating Limits Assume the following: lim f (x) = 2, lim g (x) = 1, lim h(x) = 1 x!c x!c x!c Compute the following limits: p p 3. lim 3 f (x) 4. lim g (x) x!c x!c 9 of 22 Evaluating Limits Assume the following: lim f (x) = 2, lim g (x) = 1, lim h(x) = 1 x!c x!c x!c Compute the following limits: 5. lim (h(x))2 6. lim [(h(x))3 3]2 x!c x!c Sim((x))" 3332 - [imh(x) -m = 2 = Slim , h(xs]" - my = [( -)3 - 3) = ( - 4)2 = 16 10 of 22 Evaluating Limits Use limit theorems for the following. 11 of 22 Evaluating Limits e Use limit theorems for the following. 49 Example 1: Evaluate: lim (x 3 + 2x 2 3x + 4) x! 1 lim (x3 + 2x2 - 3x + 4) = lim Climx2-Bimx + + X+ - Ii X - 2[x] + (imx] + 3) - = 2 = ( - 1)3 + 2) 1) - + 3 + 4 - 1 t 2 + 3 + 4 = = 8 11 of 22 Evaluating Limits 3x 2 + x 4 Example 2: Evaluate: lim x! 2 x 1 slimx" limx-Im Im + X+ 2 - X+ 2 - - limX- Im I X+ - 2 X+ 2 - : I ↳ - 3 I - 2 12 of 22 Evaluating Limits p 2x + 1 Example 3: Evaluate: lim x!4 2 x xim X+ 4 + lim2- In X- 4 · - 13 of 22 = Evaluating Limits Theorem Let f be a polynomial function of the form f (x) = an x n + an 1 x n 1 + an 2 x n 2 + · · · + a1 x + a0. If a is a real number, then lim f (x) = f (a). x!a 14 of 22 Evaluating Limits Theorem f (x) Let h be a rational of the form h(x) = where f and g are polynomial g (x) functions. If a is a real number, and g (a) 6= 0 then f (x) f (a) lim h(x) = lim = x!a x!a g (x) g (a) 15 of 22 Evaluating Limits Example: Evaluate the following limits. 1) lim (x3 + 2x2 - 3x + 4) = ( 1)3 + 2(1)2 - 3) 1) - + 4 X - + = - 1 + 2(1) + 3 +4 = 8 lim [(x" 4)(4x 1) + - = ((3)2 + 4)[4(3)3 17 - 2) (a + 4][4(27) 1] X>- 3 - = = 13 (107) 16 of 22 = 1311 Evaluating Limits Example: Evaluate the following limits. 3)limi lim 4) = 21]2 5 + : = -F ↓ ↓ I I 2 5 16 of 22 Evaluating Limits Example: Evaluate the following limits. 5 ). im Gif lim Notice that im 5 S lim (x" 25) X+ - = ( 5)" - 25 = 0 - 5 0 : 5 +5 hm (X + 5) - = = I - and X+ - 5 I O 16 of 22 Evaluating Limits There are functions whose limits cannot be determined immediately using the Limit Theorems we have so far. In these cases, the functions must be manipulated so that the limit, if it exists, can be calculated. 17 of 22 Evaluating Limits 0 Indeterminate Form of Type 0 f (x) If lim f (x) = 0 and lim g (x) = 0, then lim is called an indeterminate x!a x!a x!a g (x) 0 form of type. 0 18 of 22 Evaluating Limits Remarks: f (a) If f (a) = 0 and g (a) = 0, then is undefined at x = a, and not indeterminate. The g (a) f (x) f (a) term indeterminate only applies to the limit lim , and not the function value. x!a g (x) g (a) 19 of 22 Evaluating Limits Remarks: f (a) If f (a) = 0 and g (a) = 0, then is undefined at x = a, and not indeterminate. The g (a) f (x) f (a) term indeterminate only applies to the limit lim , and not the function value. x!a g (x) g (a) In computing the limit of a function as x approaches a, we are not concerned with the function value at a. 0 A limit that is indeterminate of type may exist. To find the actual value, one may find an 0 expression equivalent to the original. This is commonly done by factoring or by rationalizing (if applicable). 19 of 22 2 Evaluating Limits recall : a - b2 = (a + b)(a b) - Example: Evaluate the following limits. 1) lim : X+ - 5 lim X+ - 5 5lim- X+5 X + 5 (X-5) ; - = lim X+ - 5 - 5 - 5 = I - 10 20 of 22 Evaluating Limits Example: Evaluate the following limits. 2) xm x line scratch (1)+ 4( 1) +3 = 0 lim (x2 + 4x + 3) = - = lim X - + X+ - 13X 4 - lim (3x2 -X 4) - = 3 (1)" - ( 1) - - 4 = 0 - 1+3 = X+ - - 4 lim * X+ 3 is an indeterminate form = x+ - 13x2 X - - 4 of type8 20 of 22 Evaluating Limits Example: Evaluate the following limits. 2(3)2 7(3) +3 = 0 y - scratch : 3). lim 2 4(3) + 3 0 (3) - = Imm 20 of 22 Evaluating Limits Example: Evaluate the following limits. (-1)3 + ( 1) 1 +( + 1 = 0 (1) 1 - = - Santhi - - -.lim 4 2 1 1 1 = 0 (1) - = - lim 6 m a (b + 1)(b 1) - I lim - b+ - 1 (b + 1)(b 1) - ( - 1)2 - 1 I - -1 - 1 I 8 20 of 22 Evaluating Limits combine as a single fraction ↓ "I Example: Evaluate the following limits. I t +)*[ - = 0 (x + =x+ 20 of 22 = Evaluating Limits b a + b Example: Evaluate the following limits. a - I Im m 4) - I v recall : = im ,ex - + 3) b)(a b) a2 b2 - I (a + = I - - lim =F3 x+ + I - T I 20 of 22 Evaluating Limits (a + b)(a b) - = a2 - b2 Example: Evaluate the following limits. lim2-4 & (2)"(4) 1) O ↓ Y t > 18 - t - 18 (t 2) 16 tim (v - - 2 + 4) t 18 lim - = t ) · = ↓ g 20 of 22 Evaluating Limits Example: Evaluate the following limits. 27 + X 8) lim- x+ - 273 + " 20 of 22 Evaluating Limits Example: Evaluate the following limits. 20 of 22 Evaluating Limits Example: Evaluate the following limits. 20 of 22 Evaluating Limits Example: Evaluate the following limits. 20 of 22 Evaluating Limits Example: Evaluate the following limits. 20 of 22 References 1. Math 53 Module, Institute of Mathematics ,University of the Philippines 2. Math 21 Module, Institute of Mathematics ,University of the Philippines 3. The Calculus 7th edition Leithold, LouisAddison and Wesley Publishing Co., Inc., 1996 4. CHED Basic Calculus 21 of 22 The End Thanks for your attention! =) 22 of 22

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