Mathematics - Limits Question with Solution | TestHub

Given that

$\underset{x\to 0}{\lim }\frac{{\sin }^2\left(\pi {\left(1-{\sin }^{2}x\right)}^2\right)}{x^4}$
$=\underset{x\to 0}{\lim }\frac{{\sin }^2\left(\pi \left(1+{\sin }^{4}x-2{\sin }^{2}x\right)\right)}{x^4}$

$=\underset{x\to 0}{\lim }\frac{{\sin }^2\left(\pi -\pi \left(2{\sin }^{2}x-{\sin }^{4}x\right)\right)}{x^4}$
$=\underset{x\to 0}{\lim }\frac{{\sin }^2\left(\pi \left(2{\sin }^{2}x-{\sin }^{4}x\right)\right)}{x^4}$$=\underset{x\to 0}{\lim }{\left(\frac{\sin \left[\pi \left(2{\sin }^{2}x-{\sin }^{4}x\right)\right]}{\pi \left(2{\sin }^{2}x-{\sin }^{4}x\right)}\right)}^2·\frac{{\pi }^2{\left(2{\sin }^{2}x-{\sin }^{4}x\right)}^2}{x^4}$
When $x \to 0 \Rightarrow 2{\sin }^{2}x - {\sin }^{4}x \to 0$

Let $k = 2{\sin }^{2}x - {\sin }^{4}x$

$=\underset{k\to 0}{\lim }{\left[\frac{\sin \pi k }{\pi k}\right]}^2.\underset{x\to 0}{\lim }\frac{{\pi }^2. {\sin }^{4}x {\left(2 - {\sin }^{2}x\right)}^2}{x^4}$

We know that $\underset{x\to 0}{\lim }\frac{\sin kx}{x} = k.$

$=\frac{{\pi }^2}{{\pi }^2}.{\pi }^2.\underset{x\to 0}{\lim }{\left[\frac{\sin x }{x}\right]}^4.\underset{x\to 0}{\lim }{\left(2-{\sin }^{2}x\right)}^2$

$={\pi }^2 . 1 . {\left(2\right)}^2$

$=4{\pi }^2.$