TITLE: Calculate $\lim_{x\rightarrow0}\frac{\int_0^x\cos\frac{1}{t^{2}}dt}{x}$
QUESTION [5 upvotes]: I can't use l'Hôpital because $\lim_{x\rightarrow0}\cos\frac{1}{x^{2}}$ doesn't exist. I tried to use sandwich theorem 
$$
\frac{\int_0^x\cos\frac{1}{t^{2}}dt}{x}\leq \frac{\int_0^xdt}{x}\leq1
$$
but I didn't find a good lower bound.
Any help would be appreciated, thanks in advance.

REPLY [5 votes]: Rewrite
$$\int_0^x\cos\frac{1}{t^{2}}dt = \int_0^x\bigg(-{t^3 \over 2}\bigg)\bigg({-{2 \over t^3}}\cos\frac{1}{t^{2}}\bigg)dt$$
Now integrate this by parts, intgerating ${-{2 \over t^3}}\cos\frac{1}{t^{2}}$ and differentiating $-{t^3 \over 2}$. You get
$$-{x^3 \over 2}\sin\frac{1}{x^{2}} + \int_0^x{3t^2 \over 2} \sin\frac{1}{t^{2}}dt$$
By bounding $|\sin\frac{1}{x^{2}}|, |\sin\frac{1}{t^{2}}| \leq 1$, you get that this expression is of absolute value at most ${5 x^3 \over 6}$. So dividing by $x$ and taking the limit as $x$ goes to zero will give zero.