Calculate $\lim_{\epsilon \rightarrow 0}{{1\over {\epsilon^2}}\cdot\biggl( 1-{1\over2}\int_{-1}^{1}}\sqrt{|1-\epsilon \cdot \sin (t)|}dt \biggl)$ 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.
 A: 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}
A: 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
$$\dfrac{1}{16} \int_{-1}^1\sin^2 tdt.$$
A: 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:
$$
-\frac 1 {16}(1-\frac{\sin 2} 2 ), \qquad \sin 2 = 2 \sin 1 \cos 1.
$$
