Answer
$$
\lim _{x \rightarrow 0^{+}} x^{\sqrt x}=1
$$
Work Step by Step
$$
\lim _{x \rightarrow 0^{+}} x^{\sqrt x}
$$
Let
$$
y=x^{\sqrt x}
$$
Then
$$
\ln y= \sqrt{x} \ln x= \frac{\ln x}{x^{-1 / 2}}
$$
notice that as $ \quad x \rightarrow 0^{+} \quad $ we have $ \quad \ln x \rightarrow \infty \quad $ and $ \quad x^{-1 / 2} \rightarrow \infty $
so l’Hospital’s Rule gives
$$
\begin{aligned}
\lim _{x \rightarrow 0^{+}} \ln y & =\lim _{x \rightarrow 0^{+}} \sqrt{x} \ln x \\
& =\lim _{x \rightarrow 0^{+}} \frac{\ln x}{x^{-1 / 2}} \quad \rightarrow \frac{\infty}{\infty}\\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0^{+}} \frac{1 / x}{-\frac{1}{2} x^{-3 / 2}} \\
& =-2 \lim _{x \rightarrow 0^{+}} \sqrt{x} \\
&=0
\end{aligned}
$$
Hence, we have
$$
\begin{aligned}
\lim _{x \rightarrow 0^{+}} x^{\sqrt{x}} &=\lim _{x \rightarrow 0^{+}} e^{\ln y} \\
&=e^{0}\\
&=1
\end{aligned}
$$
