TITLE: Limit of $x^{1/x}$
QUESTION [0 upvotes]: The question is:
$\lim_\limits{x\to 0^+}x^{1/x}.$
I have used exponent laws to re-write this as:
$\lim_\limits{x\to 0^+}{e}^{\ln x/x},$ however, this does not seem to help as both $1/x$ and $\ln x$ go to $\infty$.

REPLY [1 votes]: Let $t=\frac{1}{x}$. When $x\to 0^+$, $t$ is going to $+\infty$. So, the limit is going to become:
$$\lim_{t\to+\infty}\left(\frac{1}{t}\right)^t=\lim_{t\to +\infty}e^{-\ln(t)\cdot t}=0$$
Here, we have use the fact that:
$$a^b=e^{\ln(a)\cdot b}$$
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