Find-the-value-of-lim-x-pi-2-sinx-sinx-sinx-1-sinx-lnsinx-

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 = \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