Answer
$$
\lim _{x \rightarrow 0^{+}}(4 x+1)^{\cot x}=e^{4}
$$
Work Step by Step
$$
\lim _{x \rightarrow 0^{+}}(4 x+1)^{\cot x}
$$
Let
$$
y= (4 x+1)^{\cot x}
$$
Then
$$
\ln y=\cot x \ln (4 x+1)=\frac{\ln (4 x+1)}{\tan x}
$$
notice that as $ \quad x \rightarrow 0^{+} \quad $ we have $ \quad \ln (4 x+1) \rightarrow 0 \quad $ and $ \quad \tan x \rightarrow 0 $
so l’Hospital’s Rule gives
$$
\begin{aligned}
\lim _{x \rightarrow 0^{+}} \ln y &=\lim _{x \rightarrow 0^{+}} \frac{\ln (4 x+1)}{\tan x}
\quad\quad \rightarrow \frac{0}{0}\\
&\stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0^{+}} \frac{\frac{4}{4 x+1}}{\sec ^{2} x}=4\\
\end{aligned}
$$
Hence, we have
$$
\begin{aligned}
\lim _{x \rightarrow 0^{+}}(4 x+1)^{\cot x} &=\lim _{x \rightarrow 0^{+}} e^{\ln y} \\
&=e^{4} \\
\end{aligned}
$$
