Does the limit $\lim_{t\to 0}\ln\left(1+\frac{1}{t}\right)^t$ exist? 
$$\lim_{t\to 0}\ln\left(1+\frac{1}{t}\right)^t$$

Solved.
The existence and evaluation without LHospital rule.
 A: Your function is not defined in the region to the left however, so I would interpret $\lim_{x\to0}$ to mean $\lim_{x\to0^+}$. 
If you want to define it in this region, you can choose the branch $\ln\left(-x\right)=\ln(x)+i\pi$. ($x>0$). Then
$$\lim_{x\to0^-}\left\{x\ln\left(1+\frac1x\right)\right\}=\lim_{x\to0^-}\left\{-|x|\ln\left(\frac1{|x|}-1\right)-i\pi|x|\right\}=\cdots=0$$ 
So the limit can also be shown to be $0$ from the left after defining the function in that region.
A: For sanity, replace $\frac{1}{x} = t$ and note that as $x \to 0, t \to \infty$. So our limit is
$$\lim_{t\to \infty} \frac{ \log(1+t)}{t}$$
Now make the substitution, 
$$\log(1+t) = y$$
$$\implies t = e^y -1$$
Also note that as $t\to \infty$, $y \to \infty$. Therefore our limit is: 
$$\lim_{y \to \infty} \frac{y}{e^y -1}$$
Expanding $e^y$, 
$$e^y = 1+y + \frac{y^2}{2!} + ... $$
We get, 
$$\lim_{y \to \infty} \frac{y}{y+\frac{y^2}{2!} + ... }$$
$$= \lim_{y\to\infty} \frac{1}{1+\frac{y}{2!} + ... }$$
$$=0$$
Edit: We are only concerned with $x \to 0^+$ in the original limit as the function is not defined at $x \to 0^-$.
