Do these limit problems have a limit?

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to 0^+} (1+\sin (4x))^{\cot x} & \small \color{#3D99F6} \text{A }1^\infty \text{ case, the following applies} \\ & = \exp \left(\lim_{x \to 0^+} \cot x \sin (4x)\right) & \small \color{#3D99F6} \lim_{x \to a} (f(x))^{h(x)} = \exp \left(\lim_{x \to a} h(x)(f(x)-1) \right) \\ & = \exp \left(\color{#3D99F6} \lim_{x \to 0^+} \frac {\sin (4x)}{\tan x} \right) & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies} \\ & = \exp \left(\color{#3D99F6} \lim_{x \to 0^+} \frac {4 \cos (4x)}{\sec^2 x} \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \boxed {e^4} \end{aligned}$

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Reference : $\color{#3D99F6}1^\infty$ limit (see: 2nd method )

$\displaystyle \lim _{f(x),g(x)\rightarrow 1,∞} (f(x))^{g(x)}=e^{\lim (f(x)-1)(g(x))}$

Using the same formula, we get the limit as $e^{\frac {\sin 4x}{\tan x}}=e^4$

Let $y=(1+sin(4x))^{cot(x)}$

Since direct substitution results in an indeterminate form $1^{inf}$

we can take natural log of this equation

$ln(y)=cot(x) *ln(1+sin(4x))$

$ln(y)=ln(1+sin(4x))/(1/cot(x))$

$ln(y)=ln(1+4sin(x))/tan(x)$

Since the answer is still indeterminate as 0/0

Apply L'Hospitals rule

$4cos(x)/(1+4sinx)*sec^2x$

$4cos^3(x)/(1+4sinx)$

Substituting x=0

4/1

ln(y) when x approaches 0=4

*y=e^4 ,the limit of function. *