Evaluate $\lim_{x\to 0}\left(\frac{(1+x)^{1/x}}{e}\right)^{1/x}$

Question

Evaluate

$$\lim_{x\to 0}\left(\dfrac{(1+x)^{1/x}}{e}\right)^{1/x}$$

$$$$I recognized it as the $1^\infty$ form. Thus the limit is equal to $e^t$ where $$t=\lim_{x\to 0} \left(\dfrac{(1+x)^{1/x}}{e}-1\right)\dfrac1x$$

Albeit using L'Hopital, I cannot understand how to evaluate this limit. Could somebody please help me? Many thanks!

Answer 1

Let $$y=\left(\dfrac{(1+x)^{1/x}}{e}\right)^{1/x}$$ $$\implies \log y=\frac {\log(1+x)-x}{x^2}$$ Now either use L' Hopital twice, or use Taylor expansion for $\log(1+x)$

Answer 2

If $L$ is the desired limit then we have \begin{align} \log L &= \log\left\{\lim_{x \to 0}\left(\frac{(1 + x)^{1/x}}{e}\right)^{1/x}\right\}\notag\\ &= \lim_{x \to 0}\log\left(\frac{(1 + x)^{1/x}}{e}\right)^{1/x}\text{ (via continuity of log)}\notag\\ &= \lim_{x \to 0}\frac{1}{x}\log\left(\frac{(1 + x)^{1/x}}{e}\right)\notag\\ &= \lim_{x \to 0}\frac{1}{x}\left(\log(1 + x)^{1/x} - 1\right)\notag\\ &= \lim_{x \to 0}\frac{\log (1 + x) - x}{x^{2}}\notag\\ &= \lim_{x \to 0}\dfrac{x - \dfrac{x^{2}}{2} + o(x^{2}) - x}{x^{2}}\notag\\ &= -\frac{1}{2}\notag \end{align} Hence $L = 1/\sqrt{e}$.

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Note: It is better to use general techniques (like taking logs when both base and exponent are variables) than using limit formulas for specific type of problems. In your question you are trying to use the formula that if $f(x) \to 1$ and $g(x) \to \infty$ then limit of $\{f(x)\}^{g(x)}$ is same as that of $\exp\left\{\{f(x) - 1\}g(x)\right\}$. This formula itself comes via using the technique of logarithms and as we have seen in the above solution using logs directly is more useful here.