Evaluation of $\lim_{x \rightarrow 0^+} x^{\frac{1}{x}}$ $\lim_{x \rightarrow 0^+} x^{\frac{1}{x}}$
My workout: 
Let $y$ be the answer to the limit.
\begin{align}y = \lim_{x \rightarrow 0^+} x^{\frac{1}{x}}&\implies \ln\ y = \lim_{x \rightarrow 0^+} \ln\ x^{\frac{1}{x}}\\&\implies\ln\ y = \lim_{x \rightarrow 0^+} \frac{1}{x} \ln\ x \\&\implies\ln\ y = \lim_{x \rightarrow 0^+} \frac{\ln\ x}{x}\end{align}
and by L'Hopital's Rule:
$$\ln\ y = \lim_{x \rightarrow 0^+} \frac{1}{x} \implies y = e^{\lim_{x \rightarrow 0^+} \frac{1}{x}}$$
Therefore: $y = e^{\infty} = \infty$. Correct Answer: $0$
What is wrong with my answer? And why is the answer $0$?
 A: You do good, until applying l’Hôpital to
$$
\lim_{x\to0}\frac{\ln x}{x}
$$
that cannot be done, because it's neither $0/0$ nor $\infty/\infty$. This limit is $-\infty$, so your limit is $\lim_{t\to-\infty}e^t=0$.
A: It's much easier to set $\ln x  = t, x= e^t$, to get $-\lim_{t \to \infty} t e^t = - \infty$, so the result is 0
A: rewrite as $\lim\limits_{x\to 0^+}\exp(\ln(x^{1/x}))=\lim\limits_{x\to 0^+}\exp(\frac{\ln(x)}{x})=\exp\bigg(\lim\limits_{x\to 0^+}\frac 1 x \lim\limits_{x\to 0^+} \ln(x)\bigg)$
Let $M>0$ and let $\delta =1/M.$ then $\frac{1}{x}>\frac{1}{1/M}=M$ for all $ 0<x<\delta \Longrightarrow \lim\limits_{x\to 0^+}\frac 1 x =\infty$
and $\lim\limits_{x\to 0^+}\ln(x)=-\infty$ 
$$\boxed{\color{gray}{"\exp(-\infty\cdot \infty)"}=0}$$
A: Let $n=\frac{1}{x}$
$$0\leq\lim_{x\rightarrow 0^+} x^{1/x}=\lim_{n\rightarrow \infty}\left(\frac{1}{n}\right)^n\leq\lim_{n\rightarrow \infty}\left(\frac{1}{n}\right)=0$$
A: You should've gotten that $\lim\limits_{x\to0^+} \dfrac{\ln x}{x}=-\infty$.
Proof: let $y=\ln x.$ Then $x\to 0^+\Leftrightarrow y\to -\infty.$ Thus the limit is equivalent to $\lim\limits_{y\to -\infty} y e^{-y}=-\infty.$ Since this is the limit of $\ln y$ and we want to find the limit of $y,$ we can raise $e$ to the power of the limit (we can do this because $e$ is a continuous function and $e^{\ln x} = x$.) Doing so yields $e^{-\infty} = 0.$
