Compute $ \lim_{x\to\infty}\frac{1}{x}\int_1^x\cos\frac{1}{t}\,dt $ I have a limit I can't solve. I'm studying for a test in Calculus II.
I'm asked to compute the limit:
$$
\lim_{x\to\infty}\frac{1}{x}\int_1^x\cos\frac{1}{t}\,dt
$$
It's hinted that there is no need to compute the integral itself.
Can you help me compute the limit?
 A: For all $t \ge 1$, we have $0 < \dfrac{1}{t} \le 1$. Then, since $\cos$ is a decreasing function on $[0,1]$, we have $\cos \dfrac{1}{t} \ge \cos 1$ for all $t \ge 1$. Thus, $\displaystyle\int_{1}^{x}\cos \dfrac{1}{t}\,dt \ge \int_{1}^{x}\cos 1\,dt = (x-1)\cos 1$. 
Since $\cos 1 \approx 0.54 > 0$, $\displaystyle\lim_{x \to \infty}(x-1)\cos 1 = \infty$. Therefore, $\displaystyle\lim_{x \to \infty}\int_{1}^{x}\cos \dfrac{1}{t}\,dt = \infty$, as well. 
Now that we've shown that the integral tends to infinity, we see that the limit $\displaystyle\lim_{x \to \infty}\dfrac{1}{x}\int_{1}^{x}\cos \dfrac{1}{t}\,dt$ is an indeterminate form $(\dfrac{\infty}{\infty})$. So we may use L'Hopitals Rule (as AlphaGo's deleted answer did):
$\displaystyle\lim_{x \to \infty}\dfrac{\int_{1}^{x}\cos \tfrac{1}{t}\,dt}{x} = \lim_{x \to \infty}\dfrac{\cos \tfrac{1}{x}}{1} = \lim_{x \to \infty}\cos \dfrac{1}{x} = \cos 0 = 1$.
A: Notice it suffices to calculate $ \lim _{x \rightarrow \infty} \frac{\int _1^\sqrt{x}   \cos 1/t d t}{x}  $ and $ \lim _{x \rightarrow \infty} \frac{\int _\sqrt{x}^x   \cos 1/t d t}{x}  $ 
For the first limit we use the IVT and get 
\begin{align*}
 \lim _{x \rightarrow \infty} \frac{\int _1^\sqrt{x}   \cos 1/t d t}{x} &= \lim _{x \rightarrow \infty} \frac{ (\sqrt{x}-1) \cos 1/\xi _x }{x} \rightarrow 0 \\
\end{align*}
Since $|\cos 1/\xi _x | \leq 1$
Now for the second we use again IVT and get
\begin{align*}
 \lim _{x \rightarrow \infty} \frac{\int _\sqrt{x}^x   \cos 1/t d t}{x} &= \lim _{x \rightarrow \infty} \frac{ (x-\sqrt{x}) \cos 1/\xi _x }{x}  \\
&=1
\end{align*}
Since $\xi_x  \geq \sqrt x$ which in turn means $\xi _x \rightarrow \infty$ thus $\lim _{x \rightarrow \infty} \cos 1/\xi _x =1.$ 
Therefore $$ \lim _{x \rightarrow \infty} \frac{\int _1^x   \cos 1/t d t}{x} =1 $$
A: I'm not sure what methods you are allowed to use, but as a hint:
$\lim_{x \rightarrow x_0}(fg)(x) = \left(\lim_{x \rightarrow x_0}f(x)\right)\left(\lim_{x \rightarrow x_0}g(x)\right)$
provided that the two limits on the right hand side exist. 
