Question

$\underset{x\to 0}{\lim }\left[\frac{\sin x\tan x}{x^2}\right]$, (where $x\in \left(0,\frac{\pi }{2}\right)$ and [.] denotes the greatest integer function) =

• A. 0
• B. 1
• C. $-1$
• D. 2

Answer 1

Answer: (B)

Let $f(x)=\sin x\tan x-x^2$
$f^{\prime }(x)=\cos x\cdot \tan x+\sin x\cdot {\sec }^{2}x-2x$
$\Rightarrow f^{\prime }(x)=\sin x+\sin x{\sec }^{2}x-2x$
$\Rightarrow f^{\prime \prime }(x)=\cos x+\cos x{\sec }^{2}x+2{\sec }^{2}x\sin x\tan x-2$
$\Rightarrow f^{\prime \prime }(x)=(\cos x+\sec x-2)+2{\sec }^{2}x\sin x\tan x$
Now $\cos x+\sec x-2=(\sqrt{\cos x}-\sqrt{\sec x}{)}^2$
and $2{\sec }^{2}x\tan x\cdot \sin x>0$ because $x\in \left(0,\frac{\pi }{2}\right)$
$\Rightarrow f^{\prime \prime }(x)>0\Rightarrow f^{\prime }(x)$ is M.I.
Hence $f^{\prime }(x)>f^{\prime }(0)$
$\Rightarrow f^{\prime }(x)>0$
$\Rightarrow f(x)$ is M.I.
$\Rightarrow f(x)>0\Rightarrow \sin x\tan x-x^2>0$
$\Rightarrow \frac{\sin x\tan x}{x^2}>1$
$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{\sin x\tan x}{x^2}\right]=1$