limit of $\lim_{x\to 1} x^\frac{1}{x-1}$ 
$$\lim_{x\to 1} x^\frac{1}{x-1}$$

I am trying to understand the use of $\ln$ to find the limit (I know I can define $y=x-1$ and find the limit)
what should I do after this step? $\lim_{x\to 1} \frac{1}{x-1}\cdot \ln(x)$ and how did it change the limit (as I took the ln of the limit and changed the expression) 
I can see that $\lim_{x\to 1} \frac{1}{x-1}\cdot \ln(x)=\lim_{x\to 1} \ln(x-1)' \cdot \ln(x)$
 A: It is  a standard limit that 
$$\lim_{x\to 1}\frac{\ln x}{x-1}=1$$
Actually this quotient is but the rate of change of the ln function,, starting from $x=1$:
$$\frac{\ln x}{x-1}=\frac{\ln x-\ln 1}{x-1}\to (\ln)'(1)=1\enspace \text{as}\enspace x\to 1.$$
Thus $$\lim_{x\to1}x^{\tfrac1{x-1}}=\mathrm e.$$
A: $\lim\limits_{x\to 1}\dfrac{\ln x}{x-1}=\lim\limits_{x\to 1}\dfrac{(\ln x)'}{(x-1)'}=\lim\limits_{x\to 1}\dfrac{1/x}{1}=1$
A: $$\lim_{x\to 1} x^\frac{1}{x-1}=\lim_{x\to 1} \left(1+\frac{x-1}1\right)^\frac{1}{x-1}=\lim_{y\to 0} \left(1+y\right)^\frac{1}{y}=e$$
A: Let $y = x^{1/(x-1)}$.
We wish to find $\lim\limits_{x \to 1}y$. Call this limit $L$.
Notice that $\ln(y) = \ln(x^{1/(x-1)}) = \dfrac{1}{x-1}\ln(x)$.
Because $\ln$ is continuous at $x = 1$, it follows that 
$$\lim\limits_{x \to 1}\ln(y) = \ln\left(\lim\limits_{x \to 1}y\right) = \ln(L)\text{.}$$
Furthermore, since $\ln(y) = \dfrac{\ln(x)}{x-1}$,
$$\lim\limits_{x \to 1}\ln(y) = \lim\limits_{x \to 1}\dfrac{\ln(x)}{x-1}\text{.}$$
There are several ways to evaluate this limit, but since you know L-Hospital:
$$\lim\limits_{x \to 1}\ln(y) =\lim\limits_{x \to 1}\dfrac{\ln(x)}{x-1} = \lim\limits_{x \to 1}\dfrac{1/x}{1} = 1\text{.}$$
But, as stated earlier,
$$\lim\limits_{x \to 1}\ln(y) = \ln(L)$$
so $\ln(L) = 1$, or $L = e$.
