Question

$\displaystyle\lim _{x \rightarrow 0}\left[\frac{\sin x \tan x}{x^2}\right]$, (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-2 x$
$\Rightarrow f^{\prime}(x)=\sin x+\sin x \sec ^2 x-2 x $
$\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 $
$\Rightarrow \displaystyle\lim _{x \rightarrow 0}\left[\frac{\sin x \tan x}{x^2}\right]=1$