## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises: 51

#### Answer

$$\lim_{x\to1}\Bigg(\frac{x}{x-1}-\frac{1}{\ln x}\Bigg)=\frac{1}{2}$$

#### Work Step by Step

$$A=\lim_{x\to1}\Bigg(\frac{x}{x-1}-\frac{1}{\ln x}\Bigg)$$ $$A=\lim_{x\to1}\frac{x\ln x-x+1}{(x-1)\ln x}$$ As $x\to1$, $(x\ln x-x+1)\to(1\times\ln 1-1+1)=1\times0+0=0$ and $(x-1)\ln x\to(1-1)\ln 1=0\times0=0.$ This is an indeterminate form of $\frac{0}{0}$, so we can apply L'Hospital's Rules: $$A=\lim_{x\to1}\frac{[x\ln x-x+1]'}{[(x-1)\ln x]'}$$ $$A=\lim_{x\to1}\frac{(1\times\ln x+x\times\frac{1}{x})-1+0}{(x-1)'\ln x+(x-1)(\ln x)'}$$ $$A=\lim_{x\to1}\frac{\ln x+1-1}{1\times\ln x+\frac{x-1}{x}}$$ $$A=\lim_{x\to1}\frac{\ln x}{\ln x+\frac{x-1}{x}}$$ $$A=\lim_{x\to1}\frac{\ln x}{\frac{x\ln x+x-1}{x}}$$ $$A=\lim_{x\to1}\frac{x\ln x}{x\ln x+x-1}$$ $\lim_{x\to1}(x\ln x+x-1)=1\times\ln 1+1-1=1\times0+0=0$ and $\lim_{x\to1}(x\ln x)=1\times\ln 1=1\times0=0.$ Another indeterminate form of $\frac{0}{0}$. L'Hospital's Rule would be used here: $$A=\lim_{x\to1}\frac{(x\ln x)'}{(x\ln x+x-1)'}$$ $$A=\lim_{x\to1}\frac{1\times\ln x+x\times\frac{1}{x}}{1\times\ln x+x\times\frac{1}{x}+1-0}$$ $$A=\lim_{x\to1}\frac{\ln x+1}{\ln x+1+1}$$ $$A=\lim_{x\to1}\frac{\ln x+1}{\ln x+2}$$ $$A=\frac{\ln 1+1}{\ln 1+2}$$ $$A=\frac{0+1}{0+2}=\frac{1}{2}$$

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