Problem

Find the limit. Use l'Hospital's Rule where appro…

00:55

Problem 51 Medium Difficulty

Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.

$ \displaystyle \lim_{x\to 1} \left(\frac{x}{x - 1} - \frac{1}{\ln x} \right) $

Answer

This limit has the form $\infty-\infty$
\[
\begin{aligned}
\lim _{x \rightarrow 1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right) &=\lim _{x \rightarrow 1} \frac{x \ln x-(x-1)}{(x-1) \ln x}=\lim _{x \rightarrow 1} \frac{x(1 / x)+\ln x-1}{(x-1)(1 / x)+\ln x}=\lim _{x \rightarrow 1} \frac{\ln x}{1-(1 / x)+\ln x} \\
& \stackrel{H}{=} \lim _{x \rightarrow 1} \frac{1 / x}{1 / x^{2}+1 / x} \cdot \frac{x^{2}}{x^{2}}=\lim _{x \rightarrow 1} \frac{x}{1+x}=\frac{1}{1+1}=\frac{1}{2}
\end{aligned}
\]

More Answers

03:07

Wen Z.

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

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Video Transcript

Okay. We know we end up with 00 nly plugin. Therefore, we have to apply the hospital's rule. Still got zero over zero. Therefore, we have to do this again. It's a little complicated with the complex functions. Just keep in mind that you're doing this step by step with the derivatives. We end up with 1/2.

Amrita B.

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Top Calculus 2 / BC Educators

Grace H.

Numerade Educator

Catherine R.

Missouri State University

Samuel H.

University of Nottingham

Michael J.

Idaho State University