Limit of $x^{1/x}$

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 $\infty$.

Solution 1:

Substitutions are your friend.

Let $y=\frac 1 x \implies x=\frac 1 y$.

Then $\lim\limits_{x\to 0^+} x^{1/x} = \lim\limits_{y\to \infty} (\frac 1 y)^{y} = \lim\limits_{y\to \infty} \frac 1 {y^y} = 0$.

Solution 2:

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: $$a^b=e^{\ln(a)\cdot b}$$