Vtyjkyuk

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

Calculate the next limit:

$lim_epsilon rightarrow 01over epsilon^2cdotbiggl( 1-1over2int_-1^1sqrtdt biggl)$

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

How do I find the limit?

Please help.

We have that

$$sqrt1-x=1-dfracx2-dfracx^28+o(x^2).$$

Thus

$$dfrac2-int_-1^1 sqrt1-epsilon sin t2epsilon^2=dfrac2-int_-1^1 left(1-dfracepsilon sin t2-dfracepsilon^2 sin^2 t8right)+o(epsilon^2)2epsilon^2.$$ That is

$$dfrac2-int_-1^1 sqrt1-epsilon sin t2epsilon^2=dfracdfracepsilon^28int_-1^1sin^2 tdt+o(epsilon^2)2epsilon^2.$$ Or

$$dfrac2-int_-1^1 sqrt1-epsilon sin t2epsilon^2=dfrac116 int_-1^1sin^2 tdt+o(1).$$

So the limit is

$$dfrac116 int_-1^1sin^2 tdt.$$

For $|epsilon|leq1 $ one has:
$$|1-epsilonsin(t)|=1-epsilonsin(t) forall_tin [-1,1]$$
Using L'hopital's rule twice and Leibniz's differentiation under the integral sign one gets:
beginalignlim_epsilonto 0 frac1epsilon^2 left(1-frac 1 2 int^1_-1sqrt1-epsilonsin(t),dt right)&=lim_epsilonto 0 frac12epsilon left(frac 1 2 int^1_-1fracsin(t)2sqrt1-epsilonsin(t),dt right)\
&=lim_epsilonto 0 frac12 left(frac 1 4 int^1_-1fracsin^2(t)2sqrt1-epsilonsin(t)(1-epsilonsin(t)),dt right) \
&stackrelDCT= frac116 int^1_-1sin^2(t),dt
endalign

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 $sqrt1+xsimeq 1 + x/2-x^2/8$ as $x to 0$. The integral becomes, in the limit $epsilon to 0$,
$$
int_-1^1 1-frac epsilon 2 sin t + fracepsilon^2 8 sin^2 t = 2 + fracepsilon^216 [t - sin t cos t]_-1^1=2(1+fracepsilon^216(1- sin 1 cos 1)),
$$
because $cos$ is an even function. Then it comes that your expression converges, as $epsilon to 0$, towards the following value:
$$
-frac 1 16(1-fracsin 2 2 ), qquad sin 2 = 2 sin 1 cos 1.
$$