Evaluate $\lim _{x\to \infty }\frac{1}{x}\int _0^x\cos\left(t\right)dt\:$ I think that $\lim\limits_{x\to \infty }\frac{1}{x}\int _0^x\cos\left(t\right)dt\:$ is divergent, I can prove with taylor series?
It can't be, since $\lvert \cos{t} \rvert&lt;1$, so the integral is bounded between $x$ and $-x$. It could oscillate, but not diverge. However,
$$ \frac{1}{x}\int_{0}^{x} \cos{x} \, dx = \frac{1}{x} (\sin{x}-\sin{0}) = \frac{\sin{x}}{x}, $$
which tends to $0$ since it is absolutely bounded by $1/x$, which tends to $0$.
Note that, $\int_0^x \cot (t)dt = \sin(x)$. Now, $\lim_{x \to \infty} \sin(x)/x = 0$. Therefore,   $\frac1x \int_0^x \cot (t)dt = 0.$

Note- Sorry for typing it in a hurry, I was a bit busy while I typed the solution so thats why I couldn't type in latex. I'll update it as soon as possible.