Find the limit: $\lim_\limits{x\to 0}{\frac{\left(1+x\right)^{1/x}-e}{x}}$

Find the limit: $$\lim_\limits{x\to 0}{\frac{\left(1+x\right)^{1/x}-e}{x}}$$

I have no idea what to do, but I thought that this is the limit of the derivative of $f(x)=\left(1+x\right)^{1/x}$, as $x$ tends to 0. Any help?

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If Taylor series (the “right” approach) are not yet available, let’s use L’Hospital’s Rule.

When we differentiate the top we get
The front part safely has limit $e$, so we only need to find
$$\lim_{x\to 0}\frac{x/(1+x)-\ln(1+x)}{x^2}.$$
One round of L’Hospital’s Rule gets us to
$$\lim_{x\to 0}\frac{1/(1+x)^2-1/(1+x)}{2x}.$$
Now a little algebra finishes things. Or if one likes L’Hospital’s Rule, do it again.

Indeed, it is: define the function $f$ on $\mathbb{R}$ by
f(x) = \begin{cases}(1+x)^{1/x} &\text{ if } x \neq 0 \\ e &\text{ if } x=0\end{cases}.
Then you can check that $f$ is continuous on $\mathbb{R}$, and what you want is exactly the definition of $f^\prime(0)$ (which is not guaranteed to exist: the question asks to both show its existence, and to find its value.)

To find the limit (show it exists and what its value is), I would advocate a systematic, general method for these sort of problems, namely Taylor expansions. It requires a bit of background, but once assimilated makes these limits much easier to compute:

We will need the following two expansions, where $o(\cdot)$ is the Landau notation (“little-o”), here taken when $u\to 0$:
\ln(1+u) &= u-\frac{u^2}{2} + o(u^2) \\
e^u &= 1+u + o(u)
which you can interpret as “approximations of the function on the LHS around 0 by polynomials” (here, respectively degree-2 polynomial and affine function). The $o(u^2)$ ‘means,’ in a very formal and rigorous sense, “goes to zero much faster than $u^2$, when $u\to 0$.”

  • Then, for any $x\neq 0$
    (1+x)^{1/x} = e^{\frac{1}{x}\ln(1+x)}
    and since $\ln(1+u) = u – \frac{u^2}{2} + o(u^2)$ when $u\to 0$, you get
    (1+x)^{1/x} = e^{\frac{1}{x}(x-\frac{x^2}{2}+o(x^2))} = e^{1-\frac{x}{2}+o(x)}
    = e\cdot e^{-\frac{x}{2}+o(x)}.
    Now, we use that when $u\to 0$, we have $e^u = 1+u+o(u)$: since $-\frac{x}{2}+o(x) \to 0$ when $x\to 0$, we obtain
    (1+x)^{1/x} = e\cdot e^{-\frac{x}{2}+o(x)} = e\cdot\left(1-\frac{x}{2}+o(x)\right) = e – \frac{ex}{2} + o(x).

  • It only remains to put it all together:
    \frac{(1+x)^{1/x}-e}{x} = \frac{e – \frac{ex}{2} + o(x)-e}{x}
    = – \frac{e}{2} + o(1) \xrightarrow[x\to 0]{} – \frac{e}{2}.