Question

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

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 \[x \neq \frac{\pi}{4}\]

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

Thus, the quotient function 

\[\frac{\tan \left( \frac{\pi}{4} - x \right)}{\cot 2x}\] is continuous in \[\left[ 0, \frac{\pi}{2} \right]\] for each \[x \neq \frac{\pi}{4}\] .

So, if   \[f\left( x \right)\] is continuous at 

\[x = \frac{\pi}{4}\], then it will be everywhere continuous in  \[\left[ 0, \frac{\pi}{2} \right]\] .

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

Given

\[f\left( x \right) = \frac{\tan \left( \frac{\pi}{4} - x \right)}{\cot \left( 2x \right)}, x \neq \frac{\pi}{4}\]

We have
(LHL at x = \[\frac{\pi}{4}\]) =  \[\lim_{x \to \frac{\pi}{4}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{4} - h \right) = \lim_{h \to 0} \left( \frac{\tan\left( \frac{\pi}{4} - \frac{\pi}{4} + h \right)}{\cot\left( \frac{\pi}{2} - 2h \right)} \right) = \lim_{h \to 0} \left( \frac{\tan \left( h \right)}{\tan \left( 2h \right)} \right) = \lim_{h \to 0} \left( \frac{\frac{\tan \left( h \right)}{h}}{\frac{2 \tan \left( 2h \right)}{2h}} \right) = \frac{1}{2}\left( \frac{\lim_{h \to 0} \frac{\tan \left( h \right)}{h}}{\lim_{h \to 0} \frac{\tan \left( 2h \right)}{2h}} \right) = \frac{1}{2}\]

(RHL at x = \[\frac{\pi}{4}\]) =  \[\lim_{x \to \frac{\pi}{4}^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{4} + h \right) = \lim_{h \to 0} \left( \frac{\tan \left( \frac{\pi}{4} - \frac{\pi}{4} - h \right)}{\cot \left( \frac{\pi}{2} + 2h \right)} \right) = \lim_{h \to 0} \left( \frac{\tan \left( - h \right)}{- \tan \left( 2h \right)} \right) = \lim_{h \to 0} \left( \frac{\tan \left( h \right)}{\tan \left( 2h \right)} \right) = \lim_{h \to 0} \left( \frac{\frac{\tan \left( h \right)}{h}}{\frac{2 \tan \left( 2h \right)}{2h}} \right) = \frac{1}{2}\left( \frac{\lim_{h \to 0} \frac{\tan \left( h \right)}{h}}{\lim_{h \to 0} \frac{\tan \left( 2h \right)}{2h}} \right) = \frac{1}{2}\]

If   \[f\left( x \right)\] is continuous at  \[x = \frac{\pi}{4}\] then

\[\lim_{x \to \frac{\pi}{4}^-} f\left( x \right) = \lim_{x \to \frac{\pi}{4}^+} f\left( x \right) = f\left( \frac{\pi}{4} \right)\]

∴  \[f\left( \frac{\pi}{4} \right) = \frac{1}{2}\]

Hence, for ​

\[f\left( \frac{\pi}{4} \right) = \frac{1}{2}\] , the function 

\[f\left( x \right)\] will be everywhere continuous in ​ \[\left[ 0, \frac{\pi}{2} \right]\] .