Limit of an integral in the form of $\lim_{x\to 0} g(x) \int_{0}^{x} f(t) dt$ while preparing my next exam I found this exercise in the exam of two years ago: 
$$\lim_{x\to 0} \frac{\sinh(x)}{\cos(x)-1}  \int_{0}^{x} \sqrt{e^t-t^4} dt$$
I first thought to use de L'Hopital rule, but it didn't feel right so I tried another way. I decided to try to expand the function inside the integral using McLaurin series. So the function became:
 $$\lim_{x\to 0} \frac{\sinh(x)}{\cos(x)-1}  \int_{0}^{x} (1+\frac t 2 + \frac {t^2} 4+ \frac {t^3} {12} + \frac {25t^4}{48}+\mathcal{o}(t^4)) dt$$
After expanding $\frac{\sinh(x)}{\cos(x)-1}$ and integrating and some other algebric steps, it came down to $$\frac{x^2} {\frac{-x^2} 2}$$ the result was -2.
My problem is that I'm not sure I could actually do everything I did. I someone could explain to me whether I'm right or not, and maybe also explain to me how to approach this type of exercises I would be extremely thankful. I would like to apologize already for the spelling mistakes I made for sure, but I'm not a native English speaker.
 A: What you did is right, but your first thought (L'Hospital) would be a quicker way to do this problem. If you let $$f(x)=\int_0^x\sqrt{e^x-x^4}dx,$$ then the limit is 
$$\lim_{h\rightarrow 0} \frac{\sinh (x)f(x)}{\cos (x)-1},$$
and since $f'(x)=\sqrt{e^x-x^4}$, using L'Hospital twice will give you the answer
A: Since $\sinh(x)\sim x$ and $\cos(x)-1\sim -\frac{x^2}2$ as $x\to 0$ we get
$$\frac{\sinh(x)}{\cos(x)-1}  \int_{0}^{x} \sqrt{e^t-t^4}\mathrm dt\sim-\frac2{x}\int_{0}^{x} \sqrt{e^t-t^4}\mathrm dt$$
De L'Hopital rule reduces it to
$$-2\sqrt{e^x-x^4}\xrightarrow{x\to 0}-2$$
A: The problem is easily handled by rewriting the expression under limit as $$-\dfrac{x^2} {\sin^2 x}\cdot\frac{\sinh x} {x} \cdot (\cos x +1)\cdot\frac{1}{x}\int_{0}^{x}f(t)\,dt$$ where $f(t) =\sqrt{e^t-t^4}$. Now the limit of first fraction is $1$, that of second fraction is $1$, that of third factor is $2$ and the last factor tends to $f(0)=1$ via Fundamental Theorem of Calculus. The desired answer is thus $-2$.

If you see an integral of type $\int_{0}^{x}f(t)\,dt $ in limit evaluation as $x\to 0$, it makes sense to rewrite it as $$x\cdot\frac{1}{x}\int_{0}^{x}f(t)\,dt$$ and then one uses FTC. 
