Beauty of Calculus!

Let the limit be $A$ .
Then $\displaystyle\ln A=\lim_{x\to 0}x\ln(\frac 1x)$ (from continuity of logarithm)
The right side limit $\displaystyle=\lim_{x\to 0^+}\frac{\ln(\frac 1x)}{\frac 1x}\overset{LH}=\lim_{x\to 0^+}\frac{-\frac 1x}{-\frac 1{x^2}}=0$
However, the left side limit does not exist, as $\ln \frac 1x$ is not defined for $x<0$
So the limit does not exist.

$\begin{aligned} \mathscr L & = \lim_{x \to 0} \ \left(\frac1x \right)^x \\ & = \lim_{x \to 0} e^{x\ln x} \\ & = \lim_{x \to 0} \ \color{#3D99F6}{\exp}(x\ln x) \quad \quad \small \color{#3D99F6}{\exp (x) \text{ is just another way of expressing }e^x \text{ for convenience.}} \\ & = \lim_{x \to 0} \ \exp \left( \frac{\ln x}{\frac{1}{x}} \right) \quad \quad \small \color{#3D99F6}{\text{Since } \ln x \to - \infty \text{ and }\frac 1x \to \infty \text{ as }x \to 0 \text{, we can use the L'Hôpital's rule.}} \\ & = \lim_{x \to 0} \ \exp \left( \frac{\frac{1}{x}}{-\frac{1}{x^2}} \right) \quad \quad \small \color{#3D99F6}{\text{Differentiating up and down.}} \\ & = \lim_{x \to 0} \ e^{-x} \\ & = e^0 = 1 = \boxed{0!} \end{aligned}$