TITLE: Find the limit of the following.
QUESTION [2 upvotes]: $$\lim_{x\to \ 0}\frac{{(1+x)}^\frac{1}{x}-e+ex}{x^2}$$
Using Taylor Expansion, $(1+x)^{1/x}=e(1- \frac{x}{2} + \frac{11x^2}{24} + ...)$. Finally, I found that $$\lim_{x\to \ 0}\frac{{(1+x)}^\frac{1}{x}-e+ex}{x^2}=\lim_{x\to \ 0}\frac{\frac{ex}{2}+\frac{11ex^2}{24}+...}{x^2}$$
$$=\lim_{x\to \ 0}\frac{\frac{e}{2}+\frac{11ex}{12}+...}{x}$$
Now, It seems limit doesn't exist, but options are given for this question as
1) $\frac{11e}{12}$
2)$-\frac{11e}{12}$
3)$\frac{11e}{24}$
4)$-\frac{11e}{24}$. So How to proceed further?
Thank you.

REPLY [1 votes]: You are correct in that as currently written, the limit does not exist. If the problem meant to ask $\displaystyle\lim_{x\to \ 0}\frac{{(1+x)}^\frac{1}{x}-e+\frac{ex}{2}}{x^2}$, then the answer would be $\dfrac{11e}{24}$, which is answer choice (3). My guess is that whoever wrote the question made a typo in writing $ex$ instead of $\frac{ex}{2}$ in the numerator.
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