TITLE: I'm trying to find out if this limit exists
QUESTION [6 upvotes]: I'm trying to calculate the following limit but I can't wrap my head,  around it. Can you guys give me some hints:
$$\lim_{x\to0^+}\frac{{\int_0^{x^2}\sin{\sqrt{t}}}~ dt}{x^3}$$

REPLY [19 votes]: You can also just use L'Hôpital's rule directly and the 2nd fundamental theorem of calculus: since
$$
\frac{d}{dx} \int_0^{x^2} \sin \sqrt{t} \ dt = \sin(x)\cdot2x, 
$$
the indeterminate limit is equal to
$$
\lim_{x \to 0} \frac{\sin(x)\cdot2x}{3x^2} = \frac{2}{3} \lim_{x \to 0} \frac{\sin(x)}{x}.
$$
The latter limit is known to be $1$ (from the standard proof that the derivative of the sine function is the cosine function).