TITLE: Evaluate $\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$
QUESTION [5 upvotes]: Evaluate $$\lim_{x \to \infty} \frac{1}{x} \int_x^{4x}
 \cos\left(\frac{1}{t}\right) \mbox {d}t$$

I was given the suggestion to define two functions as $g(x) = x$ and $f(x) = \int_x^{4x}\cos\left(\frac{1}{t}\right)dt$ so then if I could prove that both went to $\infty$ as $x$ went to $\infty$, then I could use L'Hôpital's rule on $\frac{f(x)}{g(x)}$; but I couldn't seem to do it for $f(x)$.
I can see that the limit is 3 if I just go ahead and differentiate both functions and take the ratio of the limits, but of course this is useless without finding my original intermediate form.
How do I show that $\frac{f(x)}{g(x)}$ is in intermediate form? or how else might I evaluate the original limit?

REPLY [2 votes]: For $x\ge\dfrac2\pi$, Dominated Convergence says
$$
\begin{align}
\lim_{x\to\infty}\frac1x\int_x^{4x}\cos\left(\frac1t\right)\,\mathrm{d}t
&=\lim_{x\to\infty}\int_1^4\cos\left(\frac1{xt}\right)\,\mathrm{d}t\\
&=\int_1^41\,\mathrm{d}t\\[9pt]
&=3
\end{align}
$$