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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 $\infty-\infty$\[\begin{aligned}\lim _{x \rightarrow 1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right) &=\lim _{x \rightarrow 1} \frac{x \ln x-(x-1)}{(x-1) \ln x}=\lim _{x \rightarrow 1} \frac{x(1 / x)+\ln x-1}{(x-1)(1 / x)+\ln x}=\lim _{x \rightarrow 1} \frac{\ln x}{1-(1 / x)+\ln x} \\& \stackrel{H}{=} \lim _{x \rightarrow 1} \frac{1 / x}{1 / x^{2}+1 / x} \cdot \frac{x^{2}}{x^{2}}=\lim _{x \rightarrow 1} \frac{x}{1+x}=\frac{1}{1+1}=\frac{1}{2}\end{aligned}\]
03:07
Wen Z.
Calculus 1 / AB
Calculus 2 / BC
Chapter 4
Applications of Differentiation
Section 4
Indeterminate Forms and l'Hospital's Rule
Derivatives
Differentiation
Volume
Campbell University
Baylor University
University of Michigan - Ann Arbor
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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.
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