Can you solve it?

$\begin{aligned} L & = \lim_{x \to 0} \left(\frac {(1 +x) ^\frac 1x} e\right) ^\frac 1{\sin x} \\ & = \lim_{x \to 0} \exp \left(\frac 1{\sin x} \left(\frac 1x\ln(1+x)-1\right) \right) \\ & = \lim_{x \to 0} \exp \left(\frac 1{\sin x} \left(\frac 1x\left (x-\frac {x^2} 2 +\frac {x^3}3-...\right) -1\right) \right) \\ & = \lim_{x \to 0} \exp \left(\frac 1{\sin x} \left(-\frac {x} 2 +\frac {x^2 }3-\frac {x^3}4...\right) \right) \\ & = \lim_{x \to 0} \exp \left(\frac 1{\frac {\sin x} x} \left(-\frac 1 2 +\frac {x}3-\frac {x^2 }4...\right) \right) \\ & = e^{-\frac 12} \end{aligned}$

$\implies k = \boxed {-\frac 12}$

In this solution, $log$ refers to the natural logarithm.

$L= \lim_{x \to 0}{\left(\dfrac{(1+x)^{\frac{1}{x}}}{e} \right )^{\frac{1}{\sin{x}}}} \\ \begin{aligned} \log{\left(L \right)} &=\lim_{x \to 0} {\left(\dfrac{\log{\left (\frac{(1+x)^{\frac{1}{x}}}{e} \right )}}{\sin{x}} \right )} \\ &=\lim_{x \to 0} {\left(\dfrac{\frac{\log{(1+x)}}{x}-\log{e}}{\sin{x}} \right )} \\ &=\lim_{x \to 0}{\left(\dfrac{\log{(1+x)}-x}{x \sin{x}} \right )} \quad \quad \color{#3D99F6}{\text{Use L'Hopital's Rule as } \dfrac{0}{0}} \\ &=\lim_{x \to 0} {\left(\dfrac{\frac{1}{1+x}-1}{\sin{x}+x \cos{x}} \right )} \quad \quad \color{#3D99F6}{\text{Use L'Hopital's Rule again as } \dfrac{0}{0}} \\ &=\lim_{x \to 0} {\left(\dfrac{-\frac{1}{(1+x)^2}}{2 \cos{x}-x \sin{x}} \right )} \\ &=-\dfrac{1}{2} \end{aligned} \\ \log{\left(L \right)}=-\dfrac{1}{2} \implies L=e^{-\frac{1}{2}} \\ \implies \color{#20A900}{\boxed{\boxed{k=-\dfrac{1}{2}}}}$