Limit calculation: $\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{e^x − 1}{x}\right)=$? For some reason I'm having trouble calculating the limit of the following function : $$\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{e^x − 1}{x}\right)$$
The function might, or might not converge. I've tried using the Euler limit but didn't get anywhere so far.
 A: Brute force via L'Hospital's rule a couple of times gives us that the limit is equal to:
$$\lim\limits_{x\rightarrow0}\frac{\frac{x}{e^x-1}\cdot\frac{xe^x-(e^x-1)}{x^2}}{1}\\
=\lim_{x\rightarrow0}\frac{xe^x-e^x+1}{xe^x-x}\\
=\lim_{x\rightarrow0}\frac{e^x+xe^x-e^x+0}{e^x+xe^x-1}\\
=\lim_{x\rightarrow0}\frac{xe^x+e^x}{2e^x+xe^x}\\
=\frac{1}{2}$$
A: You may observe that, by a Taylor expansion, you have, as $x \to 0$,
$$
\begin{align}
e^x&=1+x+\frac{x^2}2+O(x^3) \tag1
\end{align}
$$ and, as $u \to 0$, $$
\begin{align}
\ln (1+u)&=u+O(u^2) \tag2
\end{align}
$$ Thus
$$
\frac{e^x − 1}{x}=1+\frac{x}2+O(x^2) \tag3
$$ and, using $(2)$,
$$
\ln \left(\frac{e^x − 1}{x}\right)=\frac{x}2+O(x^2)
$$ giving, as $x \to 0$,

$$
\frac1{x}\ln \left(\frac{e^x − 1}{x}\right) \to \frac12.
$$

A: Notice, $$\lim_{x\to 0}\left(\frac{1}{x}\cdot \ln\frac{e^x-1}{x}\right)$$ 
 $$=\lim_{x\to 0}\frac{1}{x}\cdot \ln\left(\frac{e^x-1}{x}\right)$$ 
$$=\lim_{x\to 0}\frac{1}{x}\cdot \ln\left(\frac{1+\frac{x}{1!}+\frac{x^2}{2!}+\ldots -1}{x}\right)$$ $$=\lim_{x\to 0}\frac{1}{x}\cdot \ln\left(\frac{1}{1!}+\frac{x}{2!}+O(x^2)\right)$$ 
Now, using L-hospital's rule for $\frac{0}{0}$ form  $$=\lim_{x\to 0}\frac{\frac{d}{dx}\ln\left(\frac{1}{1!}+\frac{x}{2!}+O(x^2)\right)}{\frac{d(x)}{dx}}\cdot $$  $$=\lim_{x\to 0}\frac{\frac{d}{dx}\left(\frac{1}{1!}+\frac{x}{2!}+O(x^2)\right)}{\frac{1}{1!}+\frac{x}{2!}+O(x^2)}\cdot $$ 
$$=\lim_{x\to 0}\frac{\left(\frac{1}{2!}+O(x)\right)}{\frac{1}{1!}+\frac{x}{2!}+O(x^2)}\cdot $$  $$=\frac{\left(\frac{1}{2!}+0\right)}{\frac{1}{1!}+0}\cdot $$ $$=\frac{1}{2}$$
A: Since $(e^{x} - 1)/x \to 1$ as $x \to 0$ and further we have a $\log$ function applied to this expression, the usual technique is to express this as $$\frac{e^{x} - 1}{x} = 1 + f(x)\tag{1}$$ so that $f(x) \to 0$ as $x \to 0$. Now we need to evaluate the limit of $$\frac{\log(1 + f(x))}{x}$$ as $x \to 0$. Clearly we have
\begin{align}
L &= \lim_{x \to 0}\frac{\log(1 + f(x))}{x}\notag\\
&= \lim_{x \to 0}\frac{\log(1 + f(x))}{f(x)}\cdot\frac{f(x)}{x}\notag\\
&= \lim_{t \to 0}\frac{\log(1 + t)}{t}\cdot\lim_{x \to 0}\dfrac{\dfrac{e^{x} - 1}{x} - 1}{x}\notag\\
&= 1\cdot\lim_{x \to 0}\frac{e^{x} - 1 - x}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{e^{x} - 1}{2x}\text{ (by L'Hospital's Rule)}\notag\\
&= \frac{1}{2}\notag
\end{align}
Note that LHR is most effective when combined with the use of standard limits. Applying LHR directly in even slightly complex problems leads to complicated expressions obtained as a result of differentiation.
A: Expanding into series may help. Wolfram Alpha gives the following series expansion around $x=0$:
$$
\ln\left[ \frac{e^x-1}{x} \right] = \frac{x}{2}+\frac{x^2}{24}-\frac{x^4}{2880}+O(x^6)
$$
now dividing by $x$, the limit clearly goes to $1/2$.
