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Question Number 36018 by math1967 last updated on 27/May/18

Find the value of  lim_(x→(π/2))  ((sinx−(sinx)^(sinx) )/(1−sinx+lnsinx))

$${Find}\:{the}\:{value}\:{of} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{lim}}\:\frac{{sinx}−\left({sinx}\right)^{{sinx}} }{\mathrm{1}−{sinx}+{lnsinx}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18

t=(Π/2)−x   lim_(t→0)  ((sin((Π/2)−t)−{sin((Π/2)−t)}^({sin((Π/2)−t)}) )/(1−sin((Π/2)−t)+lnsin((Π/2)−t)))  lim_(t→0)  ((cost−(cost)^(cost) )/(1−cost+lncost))  k=cost  lim_(k→1)  ((k−k^k )/(1−k+lnk))  ((0/0))form using lh rule    y=k^k   lny=klnk  (1/y)(dy/dk)=k×(1/k)+lnk  (dy/dk)=k^k (1+lnk)  lim_(k→1)  ((1−k^k (1+lnk))/(−1+(1/k))) ((0/0))  lim_(k→1)  ((0−k^k ((1/k))−(1+lnk)(k^k )(1+lnk))/((−1)/k^2 ))  =((−1−(1+0)(1)(1+0))/(−1))  =((−2)/(−1))=2  Ans

$${t}=\frac{\Pi}{\mathrm{2}}−{x}\: \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)−\left\{{sin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)\right\}^{\left\{{sin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)\right\}} }{\mathrm{1}−{sin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)+{lnsin}\left(\frac{\Pi}{\mathrm{2}}−{t}\right)} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{cost}−\left({cost}\right)^{{cost}} }{\mathrm{1}−{cost}+{lncost}} \\ $$$${k}={cost} \\ $$$$\underset{{k}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{k}−{k}^{{k}} }{\mathrm{1}−{k}+{lnk}}\:\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right){form}\:{using}\:{lh}\:{rule} \\ $$$$ \\ $$$${y}={k}^{{k}} \\ $$$${lny}={klnk} \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dk}}={k}×\frac{\mathrm{1}}{{k}}+{lnk} \\ $$$$\frac{{dy}}{{dk}}={k}^{{k}} \left(\mathrm{1}+{lnk}\right) \\ $$$$\underset{{k}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}−{k}^{{k}} \left(\mathrm{1}+{lnk}\right)}{−\mathrm{1}+\frac{\mathrm{1}}{{k}}}\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\underset{{k}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{0}−{k}^{{k}} \left(\frac{\mathrm{1}}{{k}}\right)−\left(\mathrm{1}+{lnk}\right)\left({k}^{{k}} \right)\left(\mathrm{1}+{lnk}\right)}{\frac{−\mathrm{1}}{{k}^{\mathrm{2}} }} \\ $$$$=\frac{−\mathrm{1}−\left(\mathrm{1}+\mathrm{0}\right)\left(\mathrm{1}\right)\left(\mathrm{1}+\mathrm{0}\right)}{−\mathrm{1}} \\ $$$$=\frac{−\mathrm{2}}{−\mathrm{1}}=\mathrm{2}\:\:{Ans} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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