Question Number 36018 by math1967 last updated on 27/May/18
$${Find}\:{the}\:{value}\:{of} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{lim}}\:\frac{{sinx}−\left({sinx}\right)^{{sinx}} }{\mathrm{1}−{sinx}+{lnsinx}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18
$${t}=\frac{\Pi}{\mathrm{2}}−{x}\: \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)−\left\{{sin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)\right\}^{\left\{{sin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)\right\}} }{\mathrm{1}−{sin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)+{lnsin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{cost}−\left({cost}\right)^{{cost}} }{\mathrm{1}−{cost}+{lncost}} \\ $$$${k}={cost} \\ $$$$\underset{{k}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{k}−{k}^{{k}} }{\mathrm{1}−{k}+{lnk}}\:\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right){form}\:{using}\:{lh}\:{rule} \\ $$$$ \\ $$$${y}={k}^{{k}} \\ $$$${lny}={klnk} \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dk}}={k}×\frac{\mathrm{1}}{{k}}+{lnk} \\ $$$$\frac{{dy}}{{dk}}={k}^{{k}} \left(\mathrm{1}+{lnk}\right) \\ $$$$\underset{{k}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}−{k}^{{k}} \left(\mathrm{1}+{lnk}\right)}{−\mathrm{1}+\frac{\mathrm{1}}{{k}}}\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\underset{{k}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{0}−{k}^{{k}} \left(\frac{\mathrm{1}}{{k}}\right)−\left(\mathrm{1}+{lnk}\right)\left({k}^{{k}} \right)\left(\mathrm{1}+{lnk}\right)}{\frac{−\mathrm{1}}{{k}^{\mathrm{2}} }} \\ $$$$=\frac{−\mathrm{1}−\left(\mathrm{1}+\mathrm{0}\right)\left(\mathrm{1}\right)\left(\mathrm{1}+\mathrm{0}\right)}{−\mathrm{1}} \\ $$$$=\frac{−\mathrm{2}}{−\mathrm{1}}=\mathrm{2}\:\:{Ans} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$