please help me getting the correct answer to following problem of integral calculus

Question

Let f=n $(\sin x)^{2n+1}\cos x$ then the value of $$\lim_{n\to \infty}\int_0^{\pi/2}f dx -\int_0^{\pi/2}(\lim_{n\to \infty}f)dx$$ is (a)1/2
(b)0
(c)-1/2
(d)-$\infty$
the correct answer is (a) but i dont understand how. I evaluated $\lim_{n\to \infty}\int_0^{\pi/2}f dx$ which comes 1/2 but i did not understand how $\int_0^{\pi/2}(\lim_{n\to \infty}f)dx$ becomes "0" so to get the answer 1/2
please help.

Answer 1

$$f(x) = n(sinx)^{2n+1}cosx$$

$$\lim_{n\to \infty}\int_0^{\pi/2}f(x) dx -\int_0^{\pi/2}(\lim_{n\to \infty}f(x))dx$$

$\int_0^{\pi/2}f(x) dx$

$=\int_0^{\pi/2}n(sinx)^{2n+1}cosxdx$

$=n\int_0^{\pi/2}(sinx)^{2n+1}cosxdx$

$=n[\frac{1}{2n+2}(sinx)^{2n+2}]_{0}^{\frac{\pi}{2}}$

$=\frac{n}{2n+1}(1-0)$

$=\frac{n}{2n+1}$

Taking the limit as n goes to infinity we get $\frac{1}{2}$ as you mentioned.

$\lim_{n\to \infty}f(x)$

$=\lim_{n\to \infty}n(sinx)^{2n+1}cosx$

We check the cases $x=0$ and $x=\frac{\pi}{2}$ first.

If $x=0$,

$\lim_{n\to \infty}n(sinx)^{2n+1}cosx$

$=\lim_{n\to \infty}n(0)^{2n+1}(1)$

$=\lim_{n\to \infty}0$

$=0$

If $x=\frac{\pi}{2}$,

$\lim_{n\to \infty}n(sinx)^{2n+1}cosx$

$=\lim_{n\to \infty}n(1)^{2n+1}(0)$

$=\lim_{n\to \infty}0$

$=0$

For, $0 < x < \frac{\pi}{2}$, we know that $0 < sinx < 1$ and $0 < cosx < 1$

So in this case

$\lim_{n\to \infty}n(sinx)^{2n+1}cosx$

$=(\lim_{n\to \infty}n(sinx)^{2n+1})(\lim_{n\to \infty}cosx)$

$=cosx(\lim_{n\to \infty}n(sinx)^{2n+1})$

$=cosx(\lim_{n \to \infty}\frac{n}{(sinx)^{-2n-1}})$

numerator and denominator go to infinity so we can use l'hopital's rule.

$=cosx(\lim_{n \to \infty}\frac{1}{(-2n-1)(sinx)^{-2n-2}cosx})$

$=cosx(\lim_{n \to \infty}\frac{(sinx)^{2n+2}}{(-2n-1)cosx})$

Since we know $0 < sinx < 1$, the numerator goes to 0. The denominator goes to $-\infty$ so the limit goes to 0.

$=cosx(0)$

$=0$

So we know our limit is 0 for all cases of x.

$\int_0^{\pi/2}(\lim_{n\to \infty}f(x))dx$

$=\int_0^{\pi/2}0dx$

$=0$

So we get our final answer as $\frac{1}{2} - 0 = \frac{1}{2}$ which is a).

Answer 2

The second limit is zero since sin x hovers between 0+ and 1(-) i.e just greater than zero and just less then 1, for the interval between 0 and π/2. When n tends to infinity sin x tends to zero and cos x too hovers but anything multiplied by zero is always zero hence the limit is zero. That's why the integration itself is zero.