Question

${lim}_{x\to \frac{\pi }{4}}\frac{f\left(x\right)-f\left(\frac{\pi }{4}\right)}{x-\frac{\pi }{4}}$, where f(x) = sin 2x

Answer 1

${lim}_{x\to \frac{\pi }{4}}\left(\frac{sin2x-sin2\left(\frac{\pi }{4}\right)}{x-\frac{\pi }{4}}\right)$ [$\because$ given f(x) = sin 2x]

$={lim}_{x\to \frac{\pi }{4}}\left(\frac{sin2x-sin\frac{\pi }{2}}{x-\frac{\pi }{4}}\right)$

$\Rightarrow x\to \frac{\pi }{4}\Rightarrow x-\frac{\pi }{4}\to 0,$ let $x-\frac{\pi }{4}=y$

as $x\to \frac{\pi }{4},y\to 0$

${lim}_{y\to 0}\left(\frac{sin2(y+\frac{\pi }{4})-1}{y}\right)$

$={lim}_{y\to 0}\frac{sin(\frac{\pi }{2}+2y)-1}{y}$

$={lim}_{y\to 0}\frac{cos2y-1}{y}$

$=-{lim}_{y\to 0}\frac{1-cos2y}{y}=-{lim}_{y\to 0}\frac{2si{n}^{2}y}{y}$

$=-2{\left({lim}_{y\to 0}\frac{siny}{y}\right)}^2\times y[\because {lim}_{\theta \to 0}\frac{sin\theta }{\theta }=1]$

$=-2\times 0=0$