Mathematics - Limits Question with Solution | TestHub

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

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

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

$$\underset{x\to \frac{\pi }{2}}{\lim }f(x)=\underset{x\to \frac{\pi }{2}}{\lim }\frac{(1-\sin x)\cos x}{(2x-\pi {)}^3}\cdot \left(3{\pi }^2\right)----\left(1.62\right)$$

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

$$\begin{array}{rl}\frac{(1-\sin x)\cos x}{(2x-\pi {)}^3}& =\frac{\left[1-\sin \left(\frac{\pi +y}{2}\right)\right]\cos \left(\frac{\pi +y}{2}\right)}{y^3}\\ & =\frac{\left(1-\cos \frac{y}{2}\right)\left(-\sin \frac{y}{2}\right)}{y^3}\\ & =-\left(\frac{2{\sin }^2\frac{y}{4}}{y^2}\right)\left(\frac{\sin \frac{y}{2}}{y}\right)\\ & =-2{\left(\frac{\sin \frac{y}{4}}{y/4}\right)}^2\cdot \frac{1}{16}\cdot \left(\frac{\sin \frac{y}{2}}{y/2}\right)\cdot \frac{1}{2}\\ & =\frac{-1}{16}{\left(\frac{\sin \frac{y}{4}}{y/4}\right)}^2\left(\frac{\sin \frac{y}{2}}{y/2}\right)-----\left(1.63\right)\end{array}$$

Therefore from Eqs. (1.62) and (1.63)

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