# 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?

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$f'(x) = \cos \left( \frac{1}{4x} \right) (4x)' - \cos \left( \frac{1}{x} \right) (x)'$ check this article on wikipedia - en.wikipedia.org/wiki/Fundamental_theorem_of_calculus – qoqosz Jun 10 '12 at 18:43
I am not having trouble taking the derivative, I can evaluate the limit using L'Hôpital's just fine, but I never proved that $f(x)$ goes to $\infty$ in order to be able to use it in the first place. – stariz77 Jun 10 '12 at 18:46
stariz77 right, sorry than :) – qoqosz Jun 10 '12 at 18:47
@stariz77: That the integral blows up is clear, since $\cos(1/t)$ is nearly $1$ for large $t$. – André Nicolas Jun 10 '12 at 20:07
We are only interested in behaviour when $x$ is large. For example, let $x \ge 10$. If $t \ge x$, then $0.99 \lt \cos(1/t)\lt 1$, so $2.97 x \lt \int_x^{4x}\cos(1/t)\,dt\lt 3x$. In particular, integral has infinite limit, L'Hospital's Rule very applicable. – André Nicolas Jun 11 '12 at 3:49

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.

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How did you get the $3x\cos(\frac{1}{c})$ term? – stariz77 Jun 10 '12 at 18:56
In general we have: $$f'(c) = \frac{f(b) - f(a)}{b-a}$$ for some $c \in (a,b)$ if a<b. Now rewrite it as $(b-a) f'(c) = f(b) - f(a)$ and let $f(x) = \int^x \cos \frac{1}{t} \, dt$. In your case we have: $$(4x - x) \cos \frac{1}{c} = \int_x^{4x} \cos \frac{1}{t} \, dt$$ – qoqosz Jun 10 '12 at 19:00
@stariz77 btw by using $3x \cos \frac{1}{c}$ you can also determine that symbol for $f$ is $\infty$ :) – qoqosz Jun 10 '12 at 19:14

Hint: When $t \to + \infty$, $\cos(1/t) \to ?$

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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}

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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}

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