Question

Limit of $\underset{x\to 1}{lim}\left[\frac{{\sin }^{-1}x}{{\cot }^{-1}x}\right];$ $[.]$ is greatest integer function is

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

The correct option is B 1

Here $\underset{x\to 1}{lim}\frac{{\sin }^{-1}x}{{\cot }^{-1}x}=\frac{{lim}_{x\to 1^{-}}{\sin }^{-1}x}{{lim}_{x\to 1^{-}}{\cot }^{-1}x}$

$[\because \underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{{lim}_{x\to a}f\left(x\right)}{{lim}_{x\to a}g\left(x\right)};$ provided ${lim}_{x\to a}g\left(x\right)\ne 0]$

$=\frac{\frac{\pi }{2}}{\frac{\pi }{4}}=2$ ...(1)

Now ${\sin }^{-1}x$ is increasing and ${\cot }^{-1}x$ is a decreasing function and ${\sin }^{-1}x,{\cot }^{-1}x>0;$ when $x\to 1^{-}$

$\because \frac{f\left(x\right)}{g\left(x\right)}$ is increasing function when $f\left(x\right),g\left(x\right)>0,$ and $f\left(x\right)$ increasing and $g\left(x\right)$decreasing

$\therefore \left[\frac{{\sin }^{-1}x}{{\cot }^{-1}x}\right]<\frac{{\sin }^{-1}x}{{\cot }^{-1}x}$ where $x\to 1^{-}$

$\Rightarrow \left[\frac{{\sin }^{-1}x}{{\cot }^{-1}x}\right]<2\underset{x\to 1^{-}}{lim}\left(\frac{{\sin }^{-1}x}{{\cot }^{-1}x}\right)=2$ [from(1)]

$\Rightarrow \underset{x\to 1}{lim}\left[\frac{{\sin }^{-1}x}{{\cot }^{-1}x}\right]=1$