Compute the limit $\lim_{x\to 0}\left(\frac{(1+x)^\frac{1}{x}}{e}\right)^\frac{1}{x}$

Question

$\lim_{x\to 0}\left(\frac{(1+x)^\frac{1}{x}}{e}\right)^\frac{1}{x}$. I tried adding $1$ and subtracting $1$ so I can use $1^\infty$ case.

Answer 1

Hint. Consider the limit of the logarithm: $$\lim_{x\to 0}\ln\left(\left(\frac{(1+x)^\frac{1}{x}}{e}\right)^\frac{1}{x}\right)=\lim_{x\to 0}\ \frac{\frac{\ln(1+x)}{x}-1}{x}=\lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}= \lim_{x\to 0}\frac{\frac{1}{1+x}-1}{2x}$$ where at the last step we used L'Hopital's rule.

What is the limit $L$ as $x\to 0$? Then required limit would be $e^L$.

Answer 2

I computed the limit: $$\lim_{x\to0}\bigg(\frac{(1+x)^{1/x}}{e}\bigg)^{1/x}=\frac{1}{\sqrt{e}}.$$

Answer 3

Taking the logarithm, $$\frac{\dfrac{\log(1+x)}x-1}x=\frac{\log(1+x)-x}{x^2}=\frac{x-\frac{x^2}2+\frac{x^3}3-\cdots-x}{x^2}\to-\frac12$$

Hence

$$\frac1{\sqrt e}.$$

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As you didn't learn the Taylor expansion, use L'Hospital like Robert Z showed.