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

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}

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Catherine R.

Missouri State University

Samuel H.

University of Nottingham

Michael J.

Idaho State University

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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.

Catherine R.

Missouri State University

Samuel H.

University of Nottingham

Michael J.

Idaho State University

Lectures

Join Bootcamp