The value of lim _{x rightarrow 0^{+}} frac{c | vedclass

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(D) For $x \in (0, 1)$,the greatest integer function $[x] = 0$. Substituting this into the expression,we get: $\lim _{x \rightarrow 0^{+}} \frac{\cos

Vedclass · Accepted Answer

$\frac{\pi}{2}$

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(D) For $x \in (0, 1)$,the greatest integer function $[x] = 0$. Substituting this into the expression,we get: $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}(x-0) \cdot \sin ^{-1}(x-0)}{x(1-x^2)}$ $= \lim _{x \rightarrow 0^{+}} \left( \frac{\cos ^{-1} x}{1-x^2} \cdot \frac{\sin ^{-1} x}{x} \right)$ As $x \rightarrow 0^{+}$,$\cos ^{-1} x \rightarrow \frac{\pi}{2}$,$1-x^2 \rightarrow 1$,and $\frac{\sin ^{-1} x}{x} \rightarrow 1$. Therefore,the limit is $\frac{\pi}{2} \cdot 1 \cdot 1 = \frac{\pi}{2}$.

The value of $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(x-[x]^{2}\right) \cdot \sin ^{-1}\left(x-[x]^{2}\right)}{x-x^{3}},$ where $[x]$ denotes the greatest integer $\leq x$ is

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