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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 Hopital –  user9413 Jun 27 '11 at 12:08
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Also, not that the exponents of $x$ are a red herring. Let $y=x^2$, this becomes: $$\lim_{y \to \infty}{\frac{\int_{y}^{3y}{t\cdot \sin{\frac{2}{t}}dt}}{y}}$$ –  Thomas Andrews Jun 27 '11 at 12:38
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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 $2$. –  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
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3 Answers

up vote 3 down vote accepted

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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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... –  Shai Covo Jun 27 '11 at 14:17
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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 posted –  exTyn Jun 27 '11 at 12:19
    
Use L'hopital's rule anytime you have an ideterminate form. See math.fsu.edu/~bellenot/class/f99/cal1/indeterminate.pdf –  Nana Jun 27 '11 at 12:21
    
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
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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. –  Gerry Myerson Jun 27 '11 at 12:39
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