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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].
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Solution
When \[x \neq \frac{\pi}{4}\]
Let us consider the point x = \[\frac{\pi}{4}\] .
(LHL at x = \[\frac{\pi}{4}\]) = \[\lim_{x \to \frac{\pi}{4}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{4} - h \right) = \lim_{h \to 0} \left( \frac{\tan\left( \frac{\pi}{4} - \frac{\pi}{4} + h \right)}{\cot\left( \frac{\pi}{2} - 2h \right)} \right) = \lim_{h \to 0} \left( \frac{\tan \left( h \right)}{\tan \left( 2h \right)} \right) = \lim_{h \to 0} \left( \frac{\frac{\tan \left( h \right)}{h}}{\frac{2 \tan \left( 2h \right)}{2h}} \right) = \frac{1}{2}\left( \frac{\lim_{h \to 0} \frac{\tan \left( h \right)}{h}}{\lim_{h \to 0} \frac{\tan \left( 2h \right)}{2h}} \right) = \frac{1}{2}\]
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