Mathematics - Limits Question with Solution | TestHub

Let $f(x)=\frac{(1-\sin x)\left(8x^3-{\pi }^3\right)\cos x}{(\pi -2x{)}^4}$

$$=\frac{(1-\sin x)\cos x(2x-\pi )\left(4x^2+2\pi x+{\pi }^2\right)}{(2x-\pi {)}^4}=\frac{(1-\sin x)\cos x\left(4x^2+2\pi x+{\pi }^2\right)}{(2x-\pi {)}^3}$$

Therefore, ${\lim }_{x\to \frac{\pi }{2}}f(x)={\lim }_{x\to \frac{\pi }{2}}\frac{(1-\sin x)\cos x}{(2x-\pi {)}^3}\cdot \left(3{\pi }^2\right)$.

Put $2x-\pi =y$ so that $y\to 0$ as $x\to \pi /2$.

Therefore, now

$$\frac{(1-\sin x)\cos x}{(2x-\pi {)}^3}=\frac{\left[1-\sin \left(\frac{\pi +y}{2}\right)\right]\cos \left(\frac{\pi +y}{2}\right)}{y^3}=\frac{\left(1-\cos \frac{y}{2}\right)\left(-\sin \frac{y}{2}\right)}{y^3}$$

$$=-\left(\frac{2{\sin }^2\frac{y}{4}}{y^2}\right)\left(\frac{\sin \frac{y}{2}}{y}\right)$$