Another method for limit of $[e-(1+x)^{1/x}]/x$ as $x$ approaches zero I have solved this limit:
$\lim_{x \rightarrow 0} \frac{e-(1+x)^{\frac{1}{x}}}{x}$
using L'Hopital's rule and series expansion. Do you have other method for solving it?
 A: We can find the limit:
$$\lim_{n\to +\infty} n\left(e-\left(1+\frac{1}{n}\right)^n\right)$$
by exploiting the convexity of the exponential function, for which:
$$x<y\quad\Longrightarrow\quad e^x<\frac{e^y-e^x}{y-x}<e^y.$$
If we take $y=1$ and $x=n\log\left(1+\frac{1}{n}\right)$ (we know that $y>x$ by the Bernoulli inequality) we have:
$$\left(1+\frac{1}{n}\right)^n <\frac{e-\left(1+1/n\right)^n}{1-n\log(1+1/n)}<e$$
hence we just need to find the limit:
$$\lim_{n\to +\infty}\left(n-n^2\log(1+1/n)\right)=\frac{1}{2}$$
that follows from:
$$\log(1+1/n) = \frac{1}{n}-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)$$
to prove that the original limit equals $\frac{e}{2}$.
A: Let $f(x)=(x+1)^{1/x}$. We want to compute $\lim_{x\to 0}\frac{e-f(x)}{x}$. First,
$$\begin{align*}
\lim_{x\to 0}\ln f(x) &= \lim_{x\to 0}\frac{\ln (x+1)}{x}\\
&= \lim_{x\to 0}\frac{\ln (x+1)-\ln 1}{x}\\
&= [\ln(x)]'(1)\\
&= 1.
\end{align*}$$
Hence $\lim_{x\to 0}f(x)=e$. Therefore
$$\begin{align*}
\lim_{x\to 0}\frac{f(x)-e}{x}&= \lim_{y\to 0}f'(y)\\
&= \lim_{y\to 0} f(y)[\ln f]'(y)\\
&= e\lim_{y\to 0} \left(\frac{\ln(x+1)}{x}\right)'(y)\\
&= e\lim_{y\to 0} \frac{\frac{y}{y+1}-\ln(y+1)}{y^2}\\
\mbox{L'Hospital}&= e\lim_{y\to 0} \frac{\frac{1}{(y+1)^2}-\frac{1}{y+1}}{2y}\\
&= e\lim_{y\to 0} \frac{-y}{2y(y+1)^2}\\
&= \frac{-e}{2}.
\end{align*}$$
Hence we've shown that $\lim_{x\to 0}\frac{e-f(x)}{x}=\frac{e}{2}$ as desired.
A: An alternative approach: Recall by Taylor Series that $\ln(x+1)=x-\frac{x^2}{2}+o(x^3)$. Thus
$$\begin{align*}
(x+1)^{1/x}&= \exp\left[\frac{\ln(x+1)}{x}\right]\\
&= \exp\left[1-\frac{x}{2}+o(x^2)\right]\\
&= e\exp \left[-\frac{x}{2}+o(x^2)\right].
\end{align*}$$
Thus we compute
$$\begin{align*}
\lim_{x\to 0}\frac{e-(x+1)^{1/x}}{x}&= e\lim_{x\to 0}\frac{1-\exp \left[-\frac{x}{2}+o(x^2)\right]}{x}\\
&= -e\left(\exp \left[-\frac{x}{2}+o(x^2)\right]\right)'_{x=0}\\
&= -e\left(-\frac{x}{2}+o(x^2)\right)'_{x=0}\left(\exp \left[-\frac{x}{2}+o(x^2)\right]\right)_{x=0}\\
&=\frac{e}{2}.
\end{align*}$$
A: I doubt one can have a simple solution to this problem without using L'Hopital's Rule or infinite series. Here is one try based on the definition of $\log x$ as an integral $$\log x = \int_{1}^{x}\frac{dt}{t}$$ First we simplify the given limit $$\begin{aligned}L\,&= \lim_{x \to 0}\frac{e - (1 + x)^{1/x}}{x}\\
&= \lim_{x \to 0}\dfrac{\exp(1) - \exp\left(\dfrac{\log(1 + x)}{x}\right)}{x}\\
&= \lim_{x \to 0}\frac{e - e^{t}}{x}\text{ (putting }t = \frac{\log(1 + x)}{x}\text{ for now)}\\
&= \lim_{x \to 0}\frac{e^{t}(e^{1 - t} - 1)}{x}\\
&= \lim_{x \to 0}e^{t}\cdot\frac{e^{1 - t} - 1}{1 - t}\cdot\frac{1 - t}{x}\\
&= \lim_{t \to 1}e^{t}\cdot\lim_{t \to 1}\frac{e^{1 - t} - 1}{1 - t}\cdot\lim_{x \to 0}\frac{1 - t}{x}\\
&= e\cdot 1\cdot\lim_{x \to 0}\frac{x - \log(1 + x)}{x^{2}}\\
&= eA\end{aligned}$$ where $$A = \lim_{x \to 0}\frac{x - \log(1 + x)}{x^{2}}$$ Now to evaluate limit $A$ we need to use certain inequalities related to $\log(1 + x)$. First let $x > 0$. then we know that $$\log(1 + x) = \int_{1}^{1 + x}\frac{dt}{t} = \int_{0}^{x}\frac{dt}{1 + t}$$ Now if $0 < t < x$ we can see that $1 - t^{2} < 1 < 1 + t^{3}$ and on dividing this by $(1 + t) > 0$ we get $$1 - t < \frac{1}{1 + t} < 1 - t + t^{2}$$ Integrating this in interval $[0, x]$ we get $$x - \frac{x^{2}}{2} < \log(1 + x) < x - \frac{x^{2}}{2} + \frac{x^{3}}{3}$$ and therefore $$\frac{1}{2} - \frac{x}{3} < \frac{x - \log(1 + x)}{x^{2}} < \frac{1}{2}$$ Letting $x \to 0^{+}$ and using squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{x - \log(1 + x)}{x^{2}} = \frac{1}{2}$$ We can similarly show (using another set of inequalities) that $$\lim_{x \to 0^{-}}\frac{x - \log(1 + x)}{x^{2}} = \frac{1}{2}$$ It now follows that $A = 1/2$ and the limit $L = e/2$.
