mathop {Lim}limits_{x to {0^ - }} sin^{-1}([t | vedclass

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Question

(D) As $x 	o 0^-$,$	an x$ approaches $0$ from the negative side,i.e.,$-1 < 	an x < 0$. Therefore,$[	an x] = -1$. Substituting this into the limit:

Vedclass · Accepted Answer

$2 - \frac{\pi}{2}$

Vedclass · Answer

(D) As $x 	o 0^-$,$	an x$ approaches $0$ from the negative side,i.e.,$-1 Therefore,$[	an x] = -1$. Substituting this into the limit: $\mathop {Lim}\limits_{x 	o {0^ - }} \sin^{-1}([	an x]) = \sin^{-1}(-1) = -\frac{\pi}{2}$. So,$l = -\frac{\pi}{2}$. We need to find $\{l\} = \{-\frac{\pi}{2}\}$. Using the property $\{x\} = x - [x]$,we have $\{ -\frac{\pi}{2} \} = -\frac{\pi}{2} - [-\frac{\pi}{2}]$. Since $-1.57 Thus,$\{ -\frac{\pi}{2} \} = -\frac{\pi}{2} - (-2) = 2 - \frac{\pi}{2}$.

$\mathop {Lim}\limits_{x \to {0^ - }} \sin^{-1}([\tan x])$ $= l$,then $\{l\}$ is equal to,where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer and fractional part functions,respectively.

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