Question

Check the continuity of given function :-

$\begin{array}{ccc}f\left(x\right)=\frac{1-\sin x}{{(\frac{\pi }{2}-x)}^2},& for& x\ne \frac{\pi }{2}\\ =3,& for& x=\frac{\pi }{2}\end{array}\}$ at $x=\frac{\pi }{2}$.

Answer 1

$\underset{x\to \frac{{\pi }^{+}}{2}}{lim}\frac{1-\sin x}{(\frac{\pi }{2}-x{)}^2}=RHL$

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

$=\underset{x\to \frac{{\pi }^{+}}{2}}{lim}\frac{1-{\sin }^{2}x}{(\frac{\pi }{2}-k{)}^2(1+\sin x)}$

$=\underset{x\to \frac{{\pi }^{+}}{2}}{lim}\frac{{\cos }^{2}x}{(\frac{\pi }{2}-x{)}^2(1+\sin x)}$

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

Let $\frac{\pi }{2}-x=t$

$\Rightarrow x\to \frac{\pi }{2},t\to 0$

$RHL=\underset{t>o^{+}}{lim}\frac{{\sin }^2\to t^2}{t^2(1+8m)}$

$=\frac{1}{2}-LHL$

At $x=\frac{\pi }{2},f\left(A\right)=3$

Hence, the function $f\left(x\right)$ is not continuity, as $LHL=RHL\ne f\left(x\right)$

At $x=\frac{\pi }{2}$