Evaluate $\lim _{x\to \infty }\frac{1}{x}\int _0^x\cos\left(t\right)dt\:$

Question

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?

Answer 1

It can't be, since $\lvert \cos{t} \rvert<1$, 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 $0$.

Answer 2

Note that $$\int_0^x \cos(t)dt=\sin(x)$$ Hence, we have $$\lim_{x \to \infty} \dfrac{\displaystyle \int_0^x \cos(t)dt}x = \lim_{x \to \infty} \dfrac{\sin(x)}x = 0$$ since $\dfrac{\sin(x)}x \in \left[-\dfrac1x,\dfrac1x\right]$.

Answer 3

Such limit is zero, because $x\in\pi\mathbb{Z}$ implies $\int_{0}^{x}\cos y\,dy=0$ and for every $x\in\mathbb{R}$ we have: $$ \int_{x}^{x+\pi}|\cos x\,|\,dx \leq 2,\tag{1}$$ hence: $$ \left|\frac{1}{x}\int_{0}^{x}\cos y\,dy\right|\leq\frac{2}{x}\to 0.\tag{2}$$

Answer 4

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.