Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
$f'(x) = \cos \left( \frac{1}{4x} \right) (4x)' - \cos \left( \frac{1}{x} \right) (x)'$ check this article on wikipedia - – 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
up vote 11 down vote accepted

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.

share|cite|improve this answer
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 $t \to + \infty$, $\cos(1/t) \to ?$

share|cite|improve this answer

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

share|cite|improve this answer

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}

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.