I'm trying to evaluate the continuity and differentiability at $x = 1$ of the following piecewise function:
$$ f(x) = \begin{cases} {\Large\int_1^x \mathrm{e}^{\mathrm{-t}^\mathrm{2}}\, dt\over \Large\mathrm{(x-1)}^\alpha} -\Large{1\over 2} &\text { x > 1} \\\\ \beta &\text { x = 1} \\\\ \Large \ln x-x+1 \over \large\mathrm{(x-1)}^2 &\text { 0 < x < 1} \\ \end{cases} $$
I've been trying to compute the respective limits but I don't exactly know how to treat the first equation (maybe use l'hopital's rule along with the fundamental theorem of calculus?).
I also need to find for which values of $\alpha$ and $\beta$, $f(x)$ is continuous and differentiable.
Thank you a lot for your help.
[Edit]:
I've added the full problem I'm working on instead of just the first equation to give some more context.
Edit: This is an answer to the first version of the question, which asked just about the limit of $f$ at $1$.
Hint: Since $e^{t^2}$ is continuous, it's easy to show that $$\lim_{x\to1}\frac1{x-1}\int_1^xe^{t^2}\,dt=e^{0^2}=1.$$
Note that's not a solution, it's a hint. Meaning you can use it to find a solution.
Second hint:$$\frac{\int_1^xe^{t^2}}{(x-1)^\alpha}=(x-1)^{1-\alpha}\frac1{x-1}\int_1^xe^{t^2}.$$
This makes the answer to the following three questions obvious:
Q1: What is the limit if $\alpha<1$?
Q2: What is the limit if $\alpha=1$?
Q3: What is the limit if $\alpha>1$?