Question

The value of $\underset{x\to 0^{+}}{\lim }\frac{{\cos }^{-1}\left(x\right)·{\sin }^{-1}\left(x\right)}{x(1–x^2)}$ , where $\left[x\right]$ denotes the greatest integer $\le x$ is

• A. $0$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{2}$
• D. $\mathrm{\pi }$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{2}$

Find the value of the given expression:

Given: $\underset{x\to 0^{+}}{\lim }\frac{{\cos }^{-1}\left(x\right).{\sin }^{-1}\left(x\right)}{x(1–x^2)}$

The greatest integer function is denoted by $\left[x\right]$, for any real function.

The function rounds – off the real number down to the integer less than the number.

So, the value of $\left[x\right]$ in given expression is $0$.

Because for any number between one and zero value of $\left[x\right]$ is zero.

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\frac{{\cos }^{-1}\left(x\right).{\sin }^{-1}\left(x\right)}{x(1–x^2)} \\ \underset{x\to 0^{+}}{=\lim }\frac{{\cos }^{-1}\left(x\right)}{(1–x^2)}\times \frac{.{\sin }^{-1}\left(x\right)}{x} \\ \underset{x\to 0^{+}}{=\lim }\frac{{\cos }^{-1}\left(x\right)}{(1–x^2)}\times 1\left[\because \underset{x\to 0}{\lim }\frac{{\sin }^{-1}\left(x\right)}{x}=1\right] \\ \underset{x\to 0^{+}}{=\lim }\frac{{\cos }^{-1}\left(0^{+}\right)}{(1–{\left(0^{+}\right)}^2)} \\ =\frac{\mathrm{\pi }}{2}\left[\because {\cos }^{-1}\left(0\right)=\frac{\mathrm{\pi }}{2}\right] \end{gathered}$$

Hence option(C) i.e. $\frac{\mathrm{\pi }}{2}$ is correct.