Evaluating $\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}$

Question

How to evaluate the following limit? $$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}.$$

It depends on the tools available. My suggestion would be to rewrite $(1+x)^{1/x}$ as $\exp(\ln(1+x)/x)$ and use the Taylor series expansion for $\frac{\ln(1+x)}{x}$. — André Nicolas, Sep 05 '14 at 06:03

Olivier Oloa · Accepted Answer · 2014-09-05T19:37:12.410

2

You may write $$ \begin{align} \frac{(1+x)^{1/x} - e}{x} &= \frac{\large e^{\large\frac{\log (1+x)}{x}} - e}{x}\\\\ &= \frac{e^{\large \frac{x-\frac{x^2}{2}+{\mathcal{O}}(x^3)}{x}} - e}{x}\\\\ &= \frac{ e^{1-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\ &= \frac{ e \:e^{-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\ &= \frac{ e \:\left(1-\frac{x}{2}+{\mathcal{O}}(x^2)\right)- e}{x}\\\\ &=-\frac{e}{2}+{\mathcal{O}}(x) \end{align} $$ giving

$$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}=-\frac e2$$

where we have used $$ \begin{align} &\log(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3}+ \cdots, \quad |x|<1, \\ &e^x =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+ \cdots . \end{align} $$

edited Sep 05 '14 at 19:37

answered Sep 05 '14 at 06:10

Olivier Oloa

120,989

I don't understand how do you go from: $$\frac{e :e^{-\frac{x}{2}+o(x^2)} - e}{x}$$ to $$\frac{e :(1-\frac{x}{2}+o(x^2))- e}{x}$$. Can you elaborate a little on that? Thanks! – rubik Sep 05 '14 at 06:22
He expanded $e(\cdot)$ in Taylor series up to the second term – Alex Sep 05 '14 at 06:34
@Alex: Oh now I get it! It's because $e^x = 1 + x + o(x)$. Thank you for the tip. – rubik Sep 05 '14 at 06:40
You are welcome – Alex Sep 05 '14 at 06:50

score 1 · Answer 2 · answered Sep 05 '14 at 08:41

Since you already received very good answers, let me make the problem slightly more difficult and ask how the expression tend to $-\frac e2$.

As already answered, start with $$\frac{(1+x)^{1/x} - e}{x} = \frac{e^{\log (1+x)/x} - e}{x}$$ Now, $$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)$$ $$\frac{\log(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$ Now, define $$y=-\frac{x}{2}+\frac{x^2}{3}+\cdots$$ and $$e^{1+y}=e~ e^y=e\Big(1+y+\frac{y^2}{2}+\cdots\Big)$$ and replace $y$ by its definition from $x$. It not be too long to find that $$\frac{(1+x)^{1/x} - e}{x}=-\frac{e}{2}+\frac{11 e x}{24}+O\left(x^2\right)$$ which shows how the expression behaves when $x$ tend to $0^{\pm}$

Evaluating $\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}$

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