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$\begin{aligned} L & = \lim_{x \to 1} x^{\frac 1{x-1}} & \small \color{#3D99F6} \text{It is }1^\infty \text{ form, use } \lim_{x \to a} \left(\frac {f(x)}{g(x)} \right)^{h(x)} = e^{\lim_{x \to a} h(x) \left(\frac {f(x)}{g(x)} - 1 \right)} \\ & = e^{\lim_{x \to 1} \frac 1{x-1} (x-1)} \\ & = e \approx \boxed{2.71828182} \end{aligned}$

Let $L=\lim_{x\rightarrow 1} x^{\frac{1}{x-1}}$ . We can rewrite L using the exponential in order to make it simpler in the following way

$L=\lim_{x\rightarrow 1} x^{\frac{1}{x-1}}=\lim_{x\rightarrow 1} \exp\left(\dfrac{\ln x}{x-1}\right)$

By doing this, we get to the indeterminate form $0/0$ and it is a simple application of L'Hopital's

$L=\lim_{x\rightarrow 1} \exp\left(\dfrac{\ln x}{x-1}\right)=\lim_{x\rightarrow 1} \exp\left(\dfrac{1}{x}\right)=\exp(1)=e.$

lim x to infinity (1-x)^(1/x) = e