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The following question was taken from an exam in real analysis and functions of real variables -

Calculate the next limit:

$\lim_{\epsilon \rightarrow 0}{{1\over {\epsilon^2}}\cdot\biggl( 1-{1\over2}\int_{-1}^{1}}\sqrt{|1-\epsilon \cdot \sin (t)|}dt \biggl)$

I've tried to apply Dominant convergence theorem, but I've got messed up.

How do I find the limit?

Please help.

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  • $\begingroup$ Yes, sorry. fixed it. $\endgroup$ – Gil Or Aug 19 '18 at 10:01
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We have that

$$\sqrt{1-x}=1-\dfrac{x}{2}-\dfrac{x^2}{8}+o(x^2).$$

Thus

$$\dfrac{2-\int_{-1}^1 \sqrt{1-\epsilon \sin t}}{2\epsilon^2}=\dfrac{2-\int_{-1}^1 \left(1-\dfrac{\epsilon \sin t}{2}-\dfrac{\epsilon^2 \sin^2 t}{8}\right)+o(\epsilon^2)}{2\epsilon^2}.$$ That is

$$\dfrac{2-\int_{-1}^1 \sqrt{1-\epsilon \sin t}}{2\epsilon^2}=\dfrac{\dfrac{\epsilon^2}{8}\int_{-1}^1\sin^2 tdt+o(\epsilon^2)}{2\epsilon^2}.$$ Or

$$\dfrac{2-\int_{-1}^1 \sqrt{1-\epsilon \sin t}}{2\epsilon^2}=\dfrac{1}{16} \int_{-1}^1\sin^2 tdt+o(1).$$

So the limit is

$$\dfrac{1}{16} \int_{-1}^1\sin^2 tdt.$$

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  • $\begingroup$ Superb. I've never thought of looking at the taylor expansion. Definitely comes in handy. Thanks. $\endgroup$ – Gil Or Aug 19 '18 at 10:04
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For $|\epsilon|\leq1 $ one has: $$|1-\epsilon\sin(t)|=1-\epsilon\sin(t) \ \ \ \forall_{t\in [-1,1]}$$ Using L'hopital's rule twice and Leibniz's differentiation under the integral sign one gets: \begin{align}\lim_{\epsilon\to 0 } \frac{1}{\epsilon^2} \left(1-\frac 1 2 \int^1_{-1}\sqrt[]{1-\epsilon\sin(t)}\,dt \right)&=\lim_{\epsilon\to 0 } \frac{1}{2\epsilon} \left(\frac 1 2 \int^1_{-1}\frac{\sin(t)}{2\sqrt[]{1-\epsilon\sin(t)}}\,dt \right)\\ &=\lim_{\epsilon\to 0 } \frac{1}{2} \left(\frac 1 4 \int^1_{-1}\frac{\sin^2(t)}{2\sqrt[]{1-\epsilon\sin(t)}(1-\epsilon\sin(t))}\,dt \right) \\ &\stackrel{DCT}{=} \frac{1}{16} \int^1_{-1}\sin^2(t)\,dt \end{align}

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  • $\begingroup$ What does DCT stand for? $\endgroup$ – Zacky Aug 19 '18 at 10:35
  • $\begingroup$ @Zacky dominated convergence theorem. $\endgroup$ – Shashi Aug 19 '18 at 10:38
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There's no need for absolute values in the square root for $\epsilon \leq 1$. What i would do is to approximate the square root with $\sqrt{1+x}\simeq 1 + x/2-x^2/8$ as $x \to 0$. The integral becomes, in the limit $\epsilon \to 0$, $$ \int_{-1}^1 1-\frac \epsilon 2 \sin t + \frac{\epsilon^2} 8 \sin^2 t = 2 + \frac{\epsilon^2}{16} [t - \sin t \cos t]_{-1}^1=2(1+\frac{\epsilon^2}{16}(1- \sin 1 \cos 1)), $$ because $\cos$ is an even function. Then it comes that your expression converges, as $\epsilon \to 0$, towards the following value: $$ -\frac 1 {16}(1-\frac{\sin 2} 2 ), \qquad \sin 2 = 2 \sin 1 \cos 1. $$

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