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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.

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    Apply L Hopital2011-06-27
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    Also, not that the exponents of $x$ are a red herring. Let $y=x^2$, this becomes: $$\lim_{y \to \infty}{\frac{\int_{y}^{3y}{t\cdot \sin{\frac{2}{t}}dt}}{y}}$$2011-06-27
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    My guess would be $4$, because for $u$ small, $\sin{u}$ is approximately $u$, which, for $y$ large makes the integrand close to $2$.2011-06-27
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    @Thomas: And it's not too hard to make a proof out of your guess.2011-06-27

4 Answers 4

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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 $4$.

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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 $4$.

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    Martin's approach is simpler though...2011-06-27
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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 $4$.

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Hint:

Apply L'hopital's rule. The numerator can be differentiated using the Fundamental theorem of Calculus.

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    Does L'hopital's rule always work for such problems? I want to find general idea about solving such problems, not only this particular I posted2011-06-27
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    Use L'hopital's rule anytime you have an ideterminate form. See http://www.math.fsu.edu/~bellenot/class/f99/cal1/indeterminate.pdf2011-06-27
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    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.2011-06-27
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    @ncmath, one may note $\sin(2/t)$ is roughly $2/t$ for $t$ large, so the integrand is roughly $2$, and go from there.2011-06-27
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    @exTyn, it's good to want general ideas, but you're not going to find a general idea that works for every limit problem, not even every limit problem involving an integral. Instead, learn as many special ideas as you can, so you'll have lots of things to try on any new problem that comes up.2011-06-27
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    L'Hospital's Rule works fine here, the Marquis got a terrific deal, Bernoulli should have charged more. It is helpful but not necessary to let $x^2=u$ first.2011-06-28