Question

If $f\left(x\right)=\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x}$ for $x\ne \frac{\pi }{4},$ find the value which can be assigned to $f\left(x\right)$ at $x=\frac{\pi }{4}$ so that the function $f\left(x\right)$ become continuous every where in $[0,\pi /2]$.

Answer 1

Given,

$f\left(x\right)=\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x}$

When $x\ne \frac{\pi }{4},\tan (\frac{\pi }{4}-x)$ and $\cot 2x$ are continuous in $[0,\frac{\pi }{2}]$.

Thus, the quotient of $\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x}$ is continuous in $[0,\frac{\pi }{2}]$ for each $x\ne \frac{\pi }{4}$.

Now, for $x=\frac{\pi }{4}$, we have

LHL at $x=\frac{\pi }{4}$,

$\underset{x\to {\frac{\pi }{4}}^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(\frac{\pi }{4}-h)=\underset{h\to 0}{lim}\frac{\tan (\frac{\pi }{4}-(\frac{\pi }{4}-h))}{\cot 2(\frac{\pi }{4}-h)}=\underset{h\to 0}{lim}\frac{\tan h}{\cot (\frac{\pi }{2}-2h)}={lim}_{h\to 0}\frac{\tan h}{\tan 2h}={lim}_{h\to 0}\frac{\left(\frac{\tan h}{h}\right)}{\left(\frac{2\tan 2h}{2h}\right)}=\frac{1}{2}$

RHL at $x=\frac{\pi }{4}$,

$\underset{x\to {\frac{\pi }{4}}^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(\frac{\pi }{4}+h)=\underset{h\to 0}{lim}\frac{\tan (\frac{\pi }{4}-(\frac{\pi }{4}+h))}{\cot 2(\frac{\pi }{4}+h)}=\underset{h\to 0}{lim}\frac{\tan (-h)}{\cot (\frac{\pi }{2}+2h)}=\underset{h\to 0}{lim}\frac{-\tan h}{-\tan 2h}={lim}_{h\to 0}\frac{\left(\frac{\tan h}{h}\right)}{\left(\frac{2\tan 2h}{2h}\right)}=\frac{1}{2}$

Since $f\left(x\right)$ is continuous at $x=\frac{\pi }{4}$.

Therefore,

$\underset{x\to {\frac{\pi }{4}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{4}}^{+}}{lim}f\left(x\right)=f\left(\frac{\pi }{4}\right)$

$\Rightarrow f\left(\frac{\pi }{4}\right)=\frac{1}{2}$

Hence for $f\left(\frac{\pi }{4}\right)=\frac{1}{2},f\left(x\right)$ will be continuous everywhere in $[0,\frac{\pi }{2}]$.