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