If mathop {lim }limits_{x to 1} {sin ^{ - 1}} | vedclass

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(C) Let $L = \mathop {\lim }\limits_{x 	o 1} {\sin ^{ - 1}}\left( {k\left( {\frac{{x - 1 - \ln x}}{{(x - 1)\ln x}}} ight)} ight)$.First, evaluate

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$5$

Vedclass · Answer

(C) Let $L = \mathop {\lim }\limits_{x 	o 1} {\sin ^{ - 1}}\left( {k\left( {\frac{{x - 1 - \ln x}}{{(x - 1)\ln x}}} ight)} ight)$.First, evaluate the limit inside the $\sin^{-1}$ function:$L_{in} = k \mathop {\lim }\limits_{x 	o 1} \left( {\frac{{x - 1 - \ln x}}{{(x - 1)\ln x}}} ight)$.Let $x = 1 + h$, where $h 	o 0$:$L_{in} = k \mathop {\lim }\limits_{h 	o 0} \left( {\frac{{h - \ln(1 + h)}}{{h \ln(1 + h)}}} ight)$.Using Taylor series expansion $\ln(1 + h) = h - \frac{h^2}{2} + \frac{h^3}{3} - \dots$:$L_{in} = k \mathop {\lim }\limits_{h 	o 0} \left( {\frac{{h - (h - \frac{h^2}{2} + \dots)}}{{h(h - \frac{h^2}{2} + \dots)}}} ight) = k \mathop {\lim }\limits_{h 	o 0} \left( {\frac{{\frac{h^2}{2}}}{{h^2}}} ight) = \frac{k}{2}$.For $\sin^{-1}\left( \frac{k}{2} ight)$ to exist, the argument must satisfy $-1 \le \frac{k}{2} \le 1$.This implies $-2 \le k \le 2$.The integers in this range are $\{-2, -1, 0, 1, 2\}$.Thus, the number of integers is $5$.

If $\mathop {\lim }\limits_{x \to 1} {\sin ^{ - 1}}\left( {\frac{k}{{\ln x}} - \frac{k}{{x - 1}}} \right)$ exists, then the number of integers in the range of $k$ is:

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