Lazy Limits 1!

$\begin{aligned} \mathscr L & = \lim_{x \to 0^+} (\csc x)^\frac 1{\ln x} \\ & = \color{#3D99F6}{\exp} \left[ \ln \left( \lim_{x \to 0^+} (\csc x)^\frac 1{\ln x} \right) \right] \quad \quad \small \color{#3D99F6}{\exp (z) = e^z} \\ & = \exp \left[ \lim_{x \to 0^+} \left(\color{#3D99F6}{ - \frac {\ln (\sin x)}{\ln x}} \right) \right] \quad \quad \small \color{#3D99F6}{\text{As it is a }\infty / \infty \text{ case, L'Hôpital's rule applies.}} \\ & = \exp \left[ \lim_{x \to 0^+} \left( \color{#3D99F6}{- \frac {\frac {\cos x}{\sin x}}{\frac 1x}} \right) \right] \quad \quad \small \color{#3D99F6}{\text{Differentiating up and down w.r.t. } x} \\ & = \exp \left[ \lim_{x \to 0^+} \left(- \frac {\cos x}{\color{#3D99F6}{\frac {\sin x}x}} \right) \right] \quad \quad \small \color{#3D99F6}{x \to 0 \implies \frac {\sin x}x \to 1} \\ & = \exp \left[ - \frac 11 \right] = \boxed{e^{-1}} \end{aligned}$

$\large \displaystyle \lim_{x\to 0^{+}} (\csc x)^{\frac{1}{\ln x}} = \displaystyle e^{\ln {\lim_{x\to 0^{+}} (\csc x)^{\frac{1}{\ln x}}}} =$ due to function $\ln x$ is a continuous function $\large \displaystyle = e^{\lim_{x\to 0^{+}} \frac{1}{\ln x}\cdot \ln \csc x} =$ Applying L'Hopital's Rule $\large \displaystyle = e^{\lim_{x\to 0^{+}} \frac{x \cdot (- \cot x \cdot \csc x) }{\csc x}} = e^{\lim_{x\to 0^{+}} \frac{-x}{\tan x}} = e^{-1}$