Let f(x)=1- x4x-, x4, x(0,2).

Question

Let $f(x)=\dfrac{1-	an x}{4x-\pi}$, $x
e\dfrac\pi4$, $x\in\left(0,\dfrac\pi2ight)$. If $f(x)$ is continuous in $\left(0,\dfrac\pi2ight)$, then $f\left(\dfrac\pi4ight)$ is

Accepted Answer

$$f\left( x \right) = {{1 - \tan x} \over {4x - \pi }}$$ is continuous in $$\left[ {0,{\pi  \over 2}} \right]$$

$$\therefore$$ $$f\left( {{\pi  \over 4}} \right) = \mathop {\lim }\limits_{x \to {\pi  \over 4}} f\left( x \right) = \mathop {\lim }\limits_{x \to {\pi  \over 4}} \, + f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {{\pi  \over 4} + h} \right)$$

$$ = \mathop {\lim }\limits_{h \to 0} {{1 - \tan \left( {{\pi  \over 4} + h} \right)} \over {4\left( {{\pi  \over 4} + h} \right) - \pi }},h > 0 = \mathop {\lim }\limits_{h \to 0} {{1 - {{1 + \tan \,h} \over {1 - \tan \,h}}} \over {4h}}$$

$$ = \mathop {\lim }\limits_{h \to 0} \,{{ - 2} \over {1 - \tan \,h}}.{{\tan \,h} \over {4h}} = {{ - 2} \over 4} =  - {1 \over 2}$$

Solutions · 1