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.


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 $\alpha>1$?

  • 1
    $\begingroup$ It would be this easy if the denominator was (as you wrote) $1\over(x-1)$ but instead it is $1\over\mathrm{(x-1)}^\alpha$ so it won't just become 1 when using l'hopital's rule. $\endgroup$ – pitagorapp Jun 29 '19 at 20:56
  • $\begingroup$ @hhzi12 What do you mean, "it would be this easy..."? I didn't give a solution. Note I added another hint. $\endgroup$ – David C. Ullrich Jun 29 '19 at 23:37
  • $\begingroup$ I see now what you were trying to say. Thank you very much, the second hint made the whole problem quite trivial. $\endgroup$ – pitagorapp Jun 29 '19 at 23:59
  • $\begingroup$ @hhzi12 Possibly in the future before downvoting an answer you might consider the possibility you just didn't get it. Here for example, the second hint didn't tell you anything you didn't already know - you could have figured it out for yourself. (That's if you decided to try to think about how the first hint might help, instead of concluding I thought $(x-1)^\alpha=(x-1)$.) $\endgroup$ – David C. Ullrich Jun 30 '19 at 0:08

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