Question

Find the value of $p$ and $q$ for which
$f\left(x\right)=\{\begin{array}{ccc}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x};& if& x<\frac{x}{2}\\ p;& if& x=\frac{\pi }{2}\\ \frac{q(1-\sin x)}{(\pi -2x{)}^2};& if& x>\frac{\pi }{2}\end{array}$ is continuous at $x=\frac{\pi }{2}$.

Answer 1

left hand limit at $x=\pi /2,\underset{x\to \pi /2^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}f(\frac{\pi }{2}-h)$

$=\underset{h\to 0}{lim}\frac{1-{\sin }^3(\frac{\pi }{2}-h)}{3{\cos }^2(\frac{\pi }{2}-h)}=\frac{1-{\cos }^3(-h)}{3{\sin }^{2}h}=\frac{0}{0}$ form

By l hospital, $\underset{h\to 0}{lim}\frac{-3{\cos }^{2}h(-\sin h)}{3\times 2\sin h\cos h}=\frac{-1}{2}$

For $f\left(n\right)$ to be continuous, $p=\frac{-1}{2}$

Right hand limit at $x=\pi /2,\underset{x\to \pi /2^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f-(\frac{\pi }{2}+h)$

$=\underset{h\to 0}{lim}\frac{q(1-\sin (\frac{\pi }{2}+h))}{{(\pi -2(\frac{\pi }{2}-h))}^2}=\frac{q(1-\cos h)}{4h^2}$

As $\cos h=1-2{\sin }^{2}h/2\Rightarrow {lim}_{h\to 0}\frac{q\left(2{\sin }^{2}h/2\right)}{h^2/4.4}$

$=\underset{h\to 0}{lim}\frac{q}{4}\times 2{\left(\frac{\sin h/2}{h/2}\right)}^2=\frac{q}{2}$

$\Rightarrow \frac{q}{2}=\frac{-1}{2}\Rightarrow q=-1$