Question

$\underset{x\to \pi /2}{lim}{\tan }^{2}x[{(2{\sin }^{2}x+3\sin x+4)}^{1/2}-{({\sin }^{2}x+6\sin x+2)}^{1/2}]$ is $\frac{1}{m}$ Find $m$

Answer 1

Put $x=\frac{\pi }{2}$ in ${\tan }^{2}x[{(2{\sin }^{2}x+3\sin x+4)}^{\frac{1}{2}}-{({\sin }^{2}x+6\sin x+2)}^{\frac{1}{2}}]$

$={\tan }^2\frac{\pi }{2}[{(2{\sin }^2\frac{\pi }{2}+3\sin \frac{\pi }{2}+4)}^{\frac{1}{2}}-{({\sin }^2\frac{\pi }{2}+6\sin \frac{\pi }{2}+2)}^{\frac{1}{2}}]$

$=\infty [{(2+3+4)}^{\frac{1}{2}}-{(1+6+2)}^{\frac{1}{2}}]$

$=\infty [{\left(9\right)}^{\frac{1}{2}}-{\left(9\right)}^{\frac{1}{2}}]$

$=\infty \times 0$ is an indeterminate form

Consider ${\tan }^{2}x[{(2{\sin }^{2}x+3\sin x+4)}^{\frac{1}{2}}-{({\sin }^{2}x+6\sin x+2)}^{\frac{1}{2}}]$

$={\tan }^{2}x\left[\frac{({(2{\sin }^{2}x+3\sin x+4)}^{\frac{1}{2}}-{({\sin }^{2}x+6\sin x+2)}^{\frac{1}{2}})({(2{\sin }^{2}x+3\sin x+4)}^{\frac{1}{2}}+{({\sin }^{2}x+6\sin x+2)}^{\frac{1}{2}})}{{(2{\sin }^{2}x+3\sin x+4)}^{\frac{1}{2}}+{({\sin }^{2}x+6\sin x+2)}^{\frac{1}{2}}}\right]$

$={\tan }^{2}x\frac{2{\sin }^{2}x+3\sin x+4-{\sin }^{2}x-6\sin x-2}{\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}}$

$={\tan }^{2}x\frac{{\sin }^{2}x-3\sin x+2}{\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}}$

${lim}_{x\to \frac{\pi }{2}}{\tan }^{2}x[{(2{\sin }^{2}x+3\sin x+4)}^{\frac{1}{2}}-{({\sin }^{2}x+6\sin x+2)}^{\frac{1}{2}}]$

$={lim}_{x\to \frac{\pi }{2}}{\tan }^{2}x\frac{{\sin }^{2}x-3\sin x+2}{\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}}$

$={lim}_{x\to \frac{\pi }{2}}{\tan }^{2}x\frac{{\sin }^{2}x-3\sin x+2}{\sqrt{2{\sin }^2\frac{\pi }{2}+3\sin \frac{\pi }{2}+4}+\sqrt{{\sin }^2\frac{\pi }{2}+6\sin \frac{\pi }{2}+2}}$

$={lim}_{x\to \frac{\pi }{2}}{\tan }^{2}x\frac{{\sin }^{2}x-3\sin x+2}{\sqrt{2+3+4}+\sqrt{1+6+2}}$

$={lim}_{x\to \frac{\pi }{2}}{\tan }^{2}x\frac{{\sin }^{2}x-2\sin x-\sin x+2}{\sqrt{9}+\sqrt{9}}$

$={lim}_{x\to \frac{\pi }{2}}{\tan }^{2}x\frac{\sin x(\sin x-2)-1(\sin x-2)}{\sqrt{9}+\sqrt{9}}$

$={lim}_{x\to \frac{\pi }{2}}{\tan }^{2}x\frac{(\sin x-2)(\sin x-1)}{3+3}$

$={lim}_{x\to \frac{\pi }{2}}{\tan }^{2}x\frac{(\sin x-2)(\sin x-1)}{6}$

Put $x=\frac{\pi }{2}+h$ for $h\to 0$

$={lim}_{h\to 0}{\tan }^2(\frac{\pi }{2}+h)\frac{(\sin (\frac{\pi }{2}+h)-2)1(\sin (\frac{\pi }{2}+h)-1)}{6}$

$={lim}_{h\to 0}{\cot }^{2}h\frac{(\cos h-2)(\cos h-1)}{6}$

$={lim}_{h\to 0}\frac{(\cos h-2)(\cos h-1)}{6{\tan }^{2}h}$

$={lim}_{h\to 0}\frac{(\cos h-2)(\cos h-1)}{6h^2\frac{{\tan }^{2}h}{h^2}}$

$={lim}_{h\to 0}\frac{(\cos h-2)(\cos h-1)}{6h^2}$

By using L'Hospitals Rule we get

$={lim}_{h\to 0}\frac{(\cos h-2)(-\sin h)+(\cos h-1)(-\sin h)}{12h}$

$={lim}_{h\to 0}\frac{-\sin h(\cos h-2+\cos h-1)}{12h}$

$={lim}_{h\to 0}\frac{-\sin h(2\cos h-3)}{12h}$

$={lim}_{h\to 0}\frac{-(2\cos h-3)}{12}$

$=-\frac{2}{12}+\frac{3}{12}=\frac{1}{12}$

$\Rightarrow \frac{1}{m}=\frac{1}{12}$

$\therefore m=12$