How to calculate: $ \lim \limits_{x \to 0^+} \frac{\int_{0}^{x} (e^{t^2}-1)dt}{{\int_{0}^{x^2} \sin(t)dt}} $ How do I calculate the follwing Limit:
$$ \lim \limits_{x \to 0^+} \frac{\int_{0}^{x} (e^{t^2}-1)dt}{{\int_{0}^{x^2} \sin(t)dt}} $$
I have been solving an exam from my University's collection of previous exams, and in one of them, there wasn't an official solution, but one taken from a student's exam paper, and I couldn't understand the transition in first line.
here is a picture that I posted in Imgur ( sorry I don't have enough reputation to post pictures here so I hope u understand)

I thought about the Fundamental Theory of Calculus, because the functions integrable, then using lHopital's rule to find the limit, but I didn't get how to apply the Fundamental Thoery of Calculus here.
any kind of help would be appreciated:) !!
 A: Starting from user1337's answer, consider $$A=\frac{e^{x^2}-1}{2x\sin(x^2)}$$ when $x$ is small.
Now use Taylor expansions $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ $$\sin(y)=y-\frac{y^3}{6}+O\left(y^4\right)$$ and replace $y$ by $x^2$ so $$A=\frac{x^2+\frac{x^4}{2}+\cdots}{2x\Big(x^2-\frac{x^6}{6} +\cdots\Big)}=\frac{x^2\Big(1+\frac{x^2}{2}+\cdots\Big)}{2x^3\Big(1-\frac{x^4}{6} +\cdots\Big)}=\frac 1 {2x} \times\frac{1+\frac{x^2}{2}+\cdots}{1-\frac{x^4}{6} +\cdots}$$ When $x\to 0$, the last term goes to $1$ and then $A$ behaves as $\frac 1x$; so the limit.
A: As your limit involves the indeterminate form $\frac{0}{0}$ we examine the limit
$$\lim_{x \to 0} \frac{\frac{d}{dx} \int_0^{x} (e^{t^2}-1)dt}{\frac{d}{dx} \int_0^{x^2} \sin (t) dt}=\lim_{x \to 0} \frac{e^{x^2}-1}{2x\sin(x^2)}.$$
Where we have applied the fundamental theorem of calculus to get the RHS. Using l'Hôpital's rule a second time seems to give the answer.
A: neither the formula in the post nor in the picture makes sense. I assume you are looking for
$$ \lim \limits_{x \to 0^+} \frac{\int_{0}^{x} (e^{t^2}-1)\mathrm{d}t}{{\int_{0}^{x^2} \sin(t)\mathrm{d}t}} $$
It's obvious that both terms individually go to zero, thus, you can apply L'Hospital's rule (from your question I take, you might want to look up Leibniz' rule https://en.wikipedia.org/wiki/Leibniz_integral_rule).
A: Already typing up the solution before I saw the solution of user1337.
So from where he stopped, I continue...
$\lim_{x \to 0} \frac{e^{x^2}-1}{2x\sin(x^2)}=
\lim_{x \to 0}\frac{2xe^{x^2}}{4x^2\cos(x^2)+2\sin(x^2)}=\lim_{x \to 0}\frac{4x^2e^{x^2}+2e^{x^2}}{8x\cos(x^2)-8x^3\sin{x^2}+4x\cos(x^2)}=\frac{2}{0}$.
A: The Taylor-Maclaurin expansions $\exp(u) = 1 + u + o(u)$ and $\sin(u)=u + o(u)$ yield, as $x$ tends to $0^+$,
$$
\int_0^x (e^{t^2}-1)dt \sim\int_0^xt^2dt = \frac{x^3}{3}
$$
$$
\int_0^{x^2} \sin(t)dt\sim\int_0^{x^2}tdt = \frac{x^4}{2}
$$
Therefore
$$
\lim_{x\to0^+} \frac{\int_0^x (e^{t^2}-1)dt}{\int_0^{x^2} \sin(t)dt} = \lim_{x\to 0^+} \frac{2}{3x} = +\infty.
$$
