Answer
$$
\lim _{x \rightarrow \infty} x^{1 / x}=1
$$
Work Step by Step
$$
\lim _{x \rightarrow \infty} x^{1 / x}
$$
Let
$$
y= x^{1 / x}
$$
Then
$$
\ln y=(1 / x) \ln x =\frac{ \ln x}{x}
$$
notice that as $ \quad x \rightarrow \infty \quad $ we have $ \quad \ln x \rightarrow \infty \quad $ and $ \quad x \rightarrow \infty $
so l’Hospital’s Rule gives
$$
\begin{aligned}
\lim _{x \rightarrow \infty} \ln y &=\lim _{x \rightarrow \infty} \frac{\ln x}{x} \quad\quad \rightarrow \frac{\infty}{\infty}\\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow \infty} \frac{1 / x}{1} \\
&=0
\end{aligned}
$$
Hence, we have
$$
\begin{aligned}
\lim _{x \rightarrow \infty} x^{1 / x} &=\lim _{x \rightarrow \infty} e^{\ln y} \\
&=e^{0} \\
&=1
\end{aligned}
$$
