Limit $\lim_{x \to \infty}{\frac{\int_{x^2}^{3x^2}{t\cdot \sin{\frac{2}{t}}dt}}{x^2}}$ 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.
 A: 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 $4$.
A: 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 $4$.
A: Hint:
Apply L'hopital's rule. The numerator can be differentiated using the Fundamental theorem of Calculus.  
A: Let's put $x^{2}=u$ and let $f(t) = t\sin(2/t),f(0)=0$ and $$F(u) =\int_{0}^{u}f(t)\,dt$$ then we are seeking the limit $$\lim_{u\to\infty} \frac{F(3u)-F(u)}{u}$$ Now by mean value theorem we can see that $F(3u)-F(u)=2uf(\xi)$ for some $\xi\in(u, 3u)$ and note that as $u\to\infty$ the variable $\xi$ also tends to $ \infty$. Clearly then $f(\xi) \to 2$ and hence the desired limit is $4$.
