Evaluate $\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$ 
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? 
 A: Hint: When $t \to + \infty$, $\cos(1/t) \to ?$
A: For $x\ge\dfrac2\pi$, Dominated Convergence says
$$
\begin{align}
\lim_{x\to\infty}\frac1x\int_x^{4x}\cos\left(\frac1t\right)\,\mathrm{d}t
&=\lim_{x\to\infty}\int_1^4\cos\left(\frac1{xt}\right)\,\mathrm{d}t\\
&=\int_1^41\,\mathrm{d}t\\[9pt]
&=3
\end{align}
$$
A: For other methods of solving the limit you could use mean value theorem:
$$\frac{1}{x} \int_x^{4x} \cos \frac{1}{t} \; dt  = \frac{3x \cos \frac{1}{c}}{x}$$
for some $c \in (x,4x)$. Now when $x \to +\infty$ by squeeze theorem we get $3$ as a result.
A: Let $y = 1/t$. Then the integral becomes
\begin{align}
I & = \lim_{x \rightarrow \infty} \dfrac1x \int_x^{4x} \cos(1/t) dt = \lim_{x \rightarrow \infty} \dfrac1x \int_{1/x}^{1/(4x)} \cos(y)\dfrac{-dy}{y^2}\\
& = \lim_{x \rightarrow \infty} \dfrac1x \int_{1/(4x)}^{1/x} \dfrac{\cos(y)}{y^2} dy
\end{align}
Now use Taylor series for $\cos(y)$ and use DCT to swap limit and integral. Or equivalently, you can write $\cos(y) = 1 + \mathcal{O}(y^2)$ and then proceed.
\begin{align}
I & = \lim_{x \rightarrow \infty} \dfrac1x \int_{1/(4x)}^{1/x} \dfrac{dy}{y^2} + \lim_{x \rightarrow \infty} \dfrac1x \int_{1/(4x)}^{1/x} \mathcal{O}(1) d y = \lim_{x \rightarrow \infty} \left(\dfrac1x \left. \left( - \dfrac1y \right \rvert_{1/(4x)}^{1/x} \right) + \mathcal{O}(1/x^2) \right)\\
& = \lim_{x \rightarrow \infty} \dfrac1x \left( -x + 4x\right) = 3
\end{align}
