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


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

$F i n d t h e v a l u e o f$ $\underset{x \to \frac{π}{2}}{l i m} \frac{s i n x - {(s i n x)}^{s i n x}}{1 - s i n x + l n s i n x}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18
	$t=(Π/2)−x lim_(t→0) ((sin((Π/2)−t)−{sin((Π/2)−t)}^({sin((Π/2)−t)}) )/(1−sin((Π/2)−t)+lnsin((Π/2)−t))) lim_(t→0) ((cost−(cost)^(cost) )/(1−cost+lncost)) k=cost lim_(k→1) ((k−k^k )/(1−k+lnk)) ((0/0))form using lh rule y=k^k lny=klnk (1/y)(dy/dk)=k×(1/k)+lnk (dy/dk)=k^k (1+lnk) lim_(k→1) ((1−k^k (1+lnk))/(−1+(1/k))) ((0/0)) lim_(k→1) ((0−k^k ((1/k))−(1+lnk)(k^k )(1+lnk))/((−1)/k^2 )) =((−1−(1+0)(1)(1+0))/(−1)) =((−2)/(−1))=2 Ans$
	$t = \frac{Π}{2} - x$ $\lim_{t \to 0} \frac{s i n (\frac{Π}{2} - t) - {s i n (\frac{Π}{2} - t)}^{{s i n (\frac{Π}{2} - t)}}}{1 - s i n (\frac{Π}{2} - t) + l n s i n (\frac{Π}{2} - t)}$ $\lim_{t \to 0} \frac{c o s t - {(c o s t)}^{c o s t}}{1 - c o s t + l n c o s t}$ $k = c o s t$ $\lim_{k \to 1} \frac{k - k^{k}}{1 - k + l n k} (\frac{0}{0}) f o r m u s i n g l h r u l e$ $y = k^{k}$ $l n y = k l n k$ $\frac{1}{y} \frac{d y}{d k} = k \times \frac{1}{k} + l n k$ $\frac{d y}{d k} = k^{k} (1 + l n k)$ $\lim_{k \to 1} \frac{1 - k^{k} (1 + l n k)}{- 1 + \frac{1}{k}} (\frac{0}{0})$ $\lim_{k \to 1} \frac{0 - k^{k} (\frac{1}{k}) - (1 + l n k) (k^{k}) (1 + l n k)}{\frac{- 1}{k^{2}}}$ $= \frac{- 1 - (1 + 0) (1) (1 + 0)}{- 1}$ $= \frac{- 2}{- 1} = 2 A n s$
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