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

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

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

Therefore,

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

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

$$=-2{\left(\frac{\sin \frac{y}{4}}{y/4}\right)}^2\cdot \frac{1}{16}\cdot \left(\frac{\sin \frac{y}{2}}{y/2}\right)\cdot \frac{1}{2}=\frac{-1}{16}{\left(\frac{\sin \frac{y}{4}}{y/4}\right)}^2\left(\frac{\sin \frac{y}{2}}{y/2}\right)$$

Therefore, from Eqs. (1.62) and (1.63),

${\lim }_{x\to \frac{\pi }{2}}f(x)=\frac{-3{\pi }^2}{16}\times 1\times 1$.