Answer
$$y=e^4$$
Work Step by Step
$\lim\limits_{x \to 0^+}$ $(4x+1)^{cotx}$=
$$Let y= (4x+1)^{cotx}$$
By taking $lny$ and $ln(4x+1)^{cotx}$ and taking $\lim\limits_{x \to 0^+}$ to both
$$\lim\limits_{x \to 0^+}= \lim\limits_{x \to 0^+} cotx ln(4x+1)= 0*\infty$$
$$lny= \lim\limits_{x \to 0^+} \frac{ln(4x+1)}{tanx}$$
by using L'Hopital's Rule
$$lny= \lim\limits_{x \to 0^+} \frac{4}{\frac{4x+1}{sec^2x}}$$
$$=\lim\limits_{x \to 0^+} \frac{4sec^2x}{4x+1}$$
$$lny= \frac{4(1)}{4(0)+1}=4$$
$$y=e^4$$