Question

${lim}_{x\to \frac{\pi }{2}}\frac{\sqrt{3}-tanx}{\pi -3x}$

Answer 1

${lim}_{x\to \frac{\pi }{2}}\frac{\sqrt{3}-tanx}{\pi -3x}$

Let $x=\frac{\pi }{3}+y$

$y=x-\frac{\pi }{3}asx\to \frac{\pi }{3},y\to 0$

$={lim}_{y\to 0}\frac{\sqrt{3}-tan(\frac{\pi }{3}-y)}{3(\frac{\pi }{3}-x)}$

$={lim}_{y\to 0}\left(\frac{\sqrt{3}-\frac{tan\frac{\pi }{3}-tany}{1+tan\frac{\pi }{3}.tany}}{3y}\right)$

$={lim}_{y\to 0}\left\{\frac{(\sqrt{3}-\frac{\sqrt{3}-tany}{1+\sqrt{3}tany})}{3y}\right\}$

$={lim}_{y\to 0}\frac{(\sqrt{3}+3tany-\sqrt{3}+tany)}{3(1+\sqrt{3}tany)y}$

$={lim}_{y\to 0}\frac{4tany}{3(1+\sqrt{3}tany)}$

$=\frac{4}{3}\times {lim}_{y\to 0}\frac{tany}{y}\times \frac{1}{{lim}_{y\to 0}(1+\sqrt{3}\frac{tany}{y}\times y)}$

$=\frac{4\times 1}{3}\times \frac{1}{1+0}=\frac{4}{3}$