# $\lim_{x \to \infty}{\frac{\int_{x^2}^{3x^2}{t\cdot \sin{\frac{2}{t}}dt}}{x^2}}$

I found such problem while I was studying to my exam: $$\lim_{x \to \infty}{\frac{\int_{x^2}^{3x^2}{t\cdot \sin{\frac{2}{t}}dt}}{x^2}}$$ which I couldn't solve. I don't know methods for solving such limits with integrals.

So, what I am asking for, is a explanation and/or method(s) of solving similar limits with integrals.

Thank You.

-
Apply L Hopital –  user9413 Jun 27 '11 at 12:08
Also, not that the exponents of $x$ are a red herring. Let $y=x^2$, this becomes: $$\lim_{y \to \infty}{\frac{\int_{y}^{3y}{t\cdot \sin{\frac{2}{t}}dt}}{y}}$$ –  Thomas Andrews Jun 27 '11 at 12:38
My guess would be $4$, because for $u$ small, $\sin{u}$ is approximately $u$, which, for $y$ large makes the integrand close to $2$. –  Thomas Andrews Jun 27 '11 at 12:40
@Thomas: And it's not too hard to make a proof out of your guess. –  Hendrik Vogt Jun 27 '11 at 14:35

Since $t\cdot\sin\frac2t\to 2^-$ for $t\to\infty$, for each $\varepsilon>0$ we can choose $x_0$ such that, for $t\ge x_0$, we have $2-\varepsilon \le t\cdot\sin\frac2t \le2$.

Thus for $x\ge x_0$

$$(2-\varepsilon)2x^2 = \int_{x^2}^{3x^2} (2-\varepsilon) dt \le \int_{x^2}^{3x^2}{t\cdot \sin{\frac{2}{t}}dt} \le \int_{x^2}^{3x^2} 2 dt = 4x^2$$

and

$$(2-\varepsilon)\frac{2x^2}{x^2}=2(2-\varepsilon)\le\frac{\int_{x^2}^{3x^2}{t\cdot \sin{\frac{2}{t}}dt}}{x^2} \le \frac{4x^2}{x^2}= 4.$$

The limit is $4$.

-

Suggestion: Make a change of variable $t \mapsto t/x^2$ and recall that $\sin(u)/u \to 1$ as $u \downarrow 0$.

Elaborating:

$$\frac{{\int_{x^2 }^{3x^2 } {t\sin (\frac{2}{t})dt} }}{{x^2 }} = \frac{{\int_1^3 {x^2 t\sin (\frac{2}{{x^2 t}})x^2 dt} }}{{x^2 }} = 2\int_1^3 {\frac{{x^2 t}}{2}\sin \bigg(\frac{2}{{x^2 t}}\bigg)dt} .$$ Noting that $$\frac{{x^2 t}}{2}\sin \bigg(\frac{2}{{x^2 t}}\bigg) = \frac{{\sin (\frac{2}{{x^2 t}})}}{{2/(x^2 t)}} \to 1$$ uniformly for $t \in [1,3]$, it thus follows that the desired limit is equal to $2(3-1)$, that is to $4$.

EDIT (details concerning the uniform convergence mentioned above): Let $\varepsilon > 0$. Then $1-\varepsilon \leq \sin(u)/u \leq 1$ for all sufficiently small $u > 0$. Thus $$1 - \varepsilon \le \frac{{\sin (\frac{2}{{x^2 t}})}}{{2/(x^2 t)}} \le 1$$ for all sufficiently large $x > 0$, uniformly in $t \in [1,3]$, since $0 < 2/(x^2 t) \leq 2/x^2 \to 0$. Hence $$2\int_1^3 {(1 - \varepsilon )dt} \leq 2\int_1^3 {\frac{{x^2 t}}{2}\sin \bigg(\frac{2}{{x^2 t}}\bigg)dt} \leq 2\int_1^3 {1dt}$$ for all sufficiently large $x$, implying that the limit, as $x \to \infty$, is $4$.

-
Martin's approach is simpler though... –  Shai Covo Jun 27 '11 at 14:17
I'd be a little nervous about that. I think you need to know that the integral in the numerator grows without bound as $x\to\infty$. This is not obvious on the face of it. Some massaging of the integral may be needed to establish this. –  ncmathsadist Jun 27 '11 at 12:23
@ncmath, one may note $\sin(2/t)$ is roughly $2/t$ for $t$ large, so the integrand is roughly $2$, and go from there. –  Gerry Myerson Jun 27 '11 at 12:37