How to calculate this limit :$ \lim_{x\to 1} x^{(\frac{1}{x-1})}$ The graph of this function shows that the limit should be approximately $2.7$, but I don't know how to approach it mathematically, the form after direct putting of $x = 1$ is $1^{\infty}$, which is an indeterminate form, so we can proceed with L'-H$\hat{o}$pital's rule, but I don't really know how to do that. Please help me with this limit.
 A: Set $x-1=h$ to find $$\lim_{h\to0}(1+h)^{1/h}=\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=?$$
A: $x^{{1\over{x-1}}}$
$=e^{{{ln(x)}\over{x-1}}}$
$lim_{x\rightarrow 1}{{ln(x)}\over{x-1}}$ is the derivative of $ln(x)$ at $1$ which is $1$. So the limit is $e$.
A: Let $y = x^{(1/(x-1))}$
Take the natural log of both sides.
$ln(y) = (1/(x-1))\ln(x)$
Notice if you plug in $x = 1$ you get $0/0$, so you can apply L'Hospital's Rule.
Just take the derivative of the numerator and denominator (separately). The derivative of $\ln(x) = 1/x$ and the derivative of $x-1$ is just $1$.
So $\ln(y) = 1/x$.
Now this is an easy limit, $ln(y) = 1$. 
But that's not the limit you want. 
Apply the exponential function to both sides.
Then you see the limit is $e$.
Hope this helps!
A: Consider $$y=x^{\frac{1}{x-1}}$$ and let $x=1+t$ to make 
$$y=(1+t)^{\frac{1}{t}}\implies \log(y)=\frac 1 t \log(1+t)$$ which is simple using L'Hospital.
If you want more than the limit, use Taylor series to get
$$\log(y)=\frac 1 t \left(t-\frac{t^2}{2}+\frac{t^3}{3}+O\left(t^4\right) \right)=1-\frac{t}{2}+\frac{t^2}{3}+O\left(t^3\right)$$ Continue with Taylor using
$$y=e^{\log(y)}\implies y=e-\frac{e t}{2}+\frac{11 e t^2}{24}+O\left(t^3\right)$$ which shows the limit when $t \to 0$ and also  how it is approached.
