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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 0^+} \left(\frac{1}{x} - \frac{1}{e^x - 1} \right) $

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This limit has the form $\infty-\infty$\[\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)=\lim _{x \rightarrow 0^{+}} \frac{e^{x}-1-x}{x\left(e^{x}-1\right)} \stackrel{\text { }}{=} \lim _{x \rightarrow 0^{+}} \frac{e^{x}-1}{x e^{x}+e^{x}-1} \stackrel{\text { ? }}{=} \lim _{x \rightarrow 0^{+}} \frac{e^{x}}{x e^{x}+e^{x}+e^{x}}=\frac{1}{0+1+1}=\frac{1}{2}\]

02:02

Wen Zheng

01:33

Amrita Bhasin

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Derivatives

Differentiation

Volume

Missouri State University

Baylor University

University of Michigan - Ann Arbor

Boston College

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So for this problem here, what we want to do is Look at the function were given and take its limit as X approaches zero from the right. So the function in this case is going to be one over x minus one over E to the X -1. We can rewrite this um and take the limit as we approach from the right. So um we would end up getting 0/0. But the function that we can form from this is E to the x -1 -1 over X times E to the x -1. That's the same function that we end up getting. Um and as a result we can take the limit when we take the like we said before we get 00, but when we take the derivative, now we get E to the X -1, and then in the denominator will end up getting E to the X. Yeah, and this one plus X. E to the X. Unfortunately, plugging in zero again gives us the indeterminant form, so we'll try this again, and that's going to give us E to the X over, um to E to the X plus X, E X. Now, thankfully were able to apply the zero Limit and we end up getting 1/2 as the final answer.

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