💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!



Numerade Educator



Problem 53 Hard 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 0^+} \left(\frac{1}{x} - \frac{1}{e^x - 1} \right) $


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}


You must be signed in to discuss.

Video Transcript

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.