Question

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

Answer 1

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

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

So, if $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{4}$, then it will be everywhere continuous in $\left[0,\frac{\mathrm{\pi }}{2}\right]$.

Now,
Let us consider the point x = $\frac{\mathrm{\pi }}{4}$.

Given: $f\left(x\right)=\frac{\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}-x\right)}{\cot \left(2x\right)},x\ne \frac{\mathrm{\pi }}{4}$

We have
(LHL at x = $\frac{\mathrm{\pi }}{4}$) = $\underset{x\to {\frac{\mathrm{\pi }}{4}}^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{4}-h\right)=\underset{h\to 0}{\lim }\left(\frac{\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}-{\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}{\cot \left({\displaystyle \frac{\mathrm{\pi }}{2}}-2h\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{\tan \left(h\right)}{{\displaystyle \tan \left(2h\right)}}\right)=\underset{h\to 0}{\lim }\left(\frac{{\displaystyle \frac{\tan \left(h\right)}{h}}}{{\displaystyle \frac{2\tan \left(2h\right)}{2h}}}\right)=\frac{1}{2}\left(\frac{\underset{h\to 0}{\lim }{\displaystyle \frac{\tan \left(h\right)}{h}}}{\underset{h\to 0}{\lim }{\displaystyle \frac{\tan \left(2h\right)}{2h}}}\right)=\frac{1}{2}$

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

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{4}$, then

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

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

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