Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 98

Answer

The solution is $$\lim_{x\to1}\frac{x\ln x+\ln x-2x+2}{x^2\ln^3 x}=\frac{1}{6}.$$

Work Step by Step

To solve this limit follow the steps below. LR will stand for Apply L'Hopital's rule: $$\lim_{x\to1}\frac{x\ln x+\ln x-2x+2}{x^2\ln^3 x}=\left[\frac{1\cdot\ln 1+\ln 1-2\cdot1+2}{1^2\cdot\ln^3 1}\right]=\left[\frac{1\cdot 0+0-2+2}{1\cdot 0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to1}\frac{(x\ln x+\ln x-2x+2)'}{(x^2\ln^3 x)'}=\lim_{x\to1}\frac{(x)'\ln x+x(\ln x)'+(\ln x)'-(2x)'+(2)'}{(x^2)'\ln^3 x+x^2(\ln^3 x)'}=\lim_{x\to1}\frac{\ln x+x\cdot\frac{1}{x}+\frac{1}{x}-2}{2x\ln^3 x+x^2\cdot3\ln^2 x(\ln x)'}=\lim_{x\to1}\frac{\ln x+\frac{1}{x}-1}{2x\ln^3 x+x^2\cdot3\ln^2 x\cdot \frac{1}{x}}=\lim_{x\to 1}\frac{\ln x+\frac{1}{x}-1}{2x\ln^3 x+3x\ln^2 x}=\left[\frac{\ln 1+\frac{1}{1}-1}{2\cdot 1\cdot\ln^3 1 +3\cdot 1\ln^2 1}\right]=\left[\frac{0+1-1}{0+0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to1}\frac{(\ln x+\frac{1}{x}-1)'}{(2x\ln^3 x+3x\ln^2 x)'}=\lim_{x\to1}\frac{\frac{1}{x}-\frac{1}{x^2}}{2(x)'\ln^3 x+2x(\ln^3 x)'+3(x)'\ln^2 x+3x(\ln^2 x)'}=\lim_{x\to1}\frac{\frac{1}{x}-\frac{1}{x^2}}{2\ln^3x+2x\cdot3\ln^2 x(\ln x)'+3\ln^2 x+3x\cdot2\ln x(\ln x)'}=\lim_{x\to1}\frac{\frac{1}{x}-\frac{1}{x^2}}{2\ln^3 x+6x\ln^2 x\cdot\frac{1}{x}+3\ln^2 x+6x\ln x\cdot \frac{1}{x}}=\lim_{x\to1}\frac{\frac{1}{x}-\frac{1}{x^2}}{2\ln^3 x+6\ln^2 x+3\ln^2 x+6\ln x}=\lim_{x\to1}\frac{\frac{1}{x}-\frac{1}{x^2}}{2\ln^3 x+9\ln^2 x+6\ln x}=\lim_{x\to1}\frac{x-1}{2x^2\ln^3 x+9x^2\ln^2 x+6x^2\ln x}=\left[\frac{1-1}{2\cdot1^2\ln^3 1+9\cdot1^2\ln^2 1+6\cdot1^2\ln 1}\right]=\left[\frac{0}{2\cdot0^3+9\cdot 0^2+6\cdot 0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to1}\frac{(x-1)'}{(x^2(2\ln^3 x+9\ln^2 x+6\ln x))'}=\lim_{x\to1}\frac{1}{x^2(2\ln^3 x+9\ln^2 x+6\ln x)'+(x^2)'(2\ln^3 x+9\ln^2 x+6\ln x)}=\lim_{x\to1}\frac{1}{x^2(2\cdot 3\ln^2 x(\ln x)'+9\cdot2\ln x(\ln x)'+6\frac{1}{x})+2x(2\ln^3 x+9\ln^2 x+6\ln x)}=\lim_{x\to1}\frac{1}{x^2\left(6\ln^2 x\cdot\frac{1}{x}+18\ln x\cdot\frac{1}{x}+6\frac{1}{x}\right)+2x(2\ln^3 x+9\ln^2 x+6\ln x)}=\lim_{x\to1}\frac{1}{6x\ln^2 x+18x\ln x+6x+4x\ln^3 x+18x\ln^2 x+12x\ln x} = \lim_{x\to1}\frac{1}{4x\ln^3+24x\ln^2 x+30x\ln x+6x}=\left[\frac{1}{0+0+0+6}\right]=\frac{1}{6}.$$
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