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Find the following limit. 5 In(x) lim x-1 X-1 Solution Since lim 5 In(x) = 5 In(1) = x-1 the limit is an indeterminate form, so we can apply l'Hospital's Rule. 5 In(x) lim X→1 X-1 = lim X→ 1 = lim = x → 1 = lim X→ 1 (5 -(5 In(x)) (x-1) 1 and lim (x - 1) = X→ 1

Find the following limit. 5 In(x) lim x-1 X-1 Solution Since lim 5 In(x) = 5 In(1) = x-1 the limit is an indeterminate form, so we can apply l'Hospital's Rule. 5 In(x) lim X→1 X-1 = lim X→ 1 = lim = x → 1 = lim X→ 1 (5 -(5 In(x)) (x-1) 1 and lim (x - 1) = X→ 1

Trigonometry (11th Edition)
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Find the following limit.
lim x → 1
 
 
 
Find the following limit. 5 In(x) lim x 1 x 1 Solution Since lim 5 In(x) = 5 In(1) = x → 1 5 In(x) lim x 1 x 1 the limit is an indeterminate form, so we can apply l'Hospital's Rule. = = lim X→ 1 = lim x → 1 = lim x → 1 d (5 In(x)) dx d -(x - 1) dx and 1 lim (x - 1) = X→ 1
Transcribed Image Text:Find the following limit. 5 In(x) lim x 1 x 1 Solution Since lim 5 In(x) = 5 In(1) = x → 1 5 In(x) lim x 1 x 1 the limit is an indeterminate form, so we can apply l'Hospital's Rule. = = lim X→ 1 = lim x → 1 = lim x → 1 d (5 In(x)) dx d -(x - 1) dx and 1 lim (x - 1) = X→ 1
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Step 1: Given Limit

Given Limit 

limit as x rightwards arrow 1 of fraction numerator 5 ln left parenthesis space x space right parenthesis over denominator x space minus space 1 end fraction

L - Hospital Rule will be applicable when limit is indeterminate form 

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