limits

\lim_{x ---> 0 } \frac{xcos x - sin x }{x ^2 sinx }

this of form o/o

use l' hospital
\lim_{x ---> 0 } \frac{- x sin x }{x ^2 cos x+ sin x . 2x }

remove outside x and take common

again differntiate
\lim_{x ---> 0 } \frac{- cos x }{x . - sin x + cosx + 2cos x } = \frac{-1}{0+ 1 + 2} = \frac{-1}{3}

pahle ke liye ans is 2/3

\lim_{x ---> 0 } \frac{[sin (x/2) + cos (x/2)]^{2/3}- [sin (x/2) + cos (x/2)]^{2/3} }{x}

yahaan bhi l' hospital use karo

on differantiation we get

\lim_{x ---> 0 } \frac{ \frac{2}{3}.[sin(x/2) + cos (x/2)]^{-1/3}[ 1/2 cos x/2 - 1/2sin x/2 ] -\frac{2}{3} .[sin x/2- cos x/2]^{-1/3}[ 1/2 cos x/2 + 1/2 sin x/2 ] }{1}

now putting x---> o we get

\lim_{x ---> 0 }\frac{2/3 .(-1 ) ^{-1/3}.[ 1/2 . 1] - 2/3 ( -1 ) ^{-1/3} . 1/2}{1}

= \frac{2}{3} .\frac{1}{2} + \frac{2}{3}.\frac{1}{2} = 2/3

note --1 ( -1 ) ^ -1/3 = 1/ -1 = -1

how to solve the 1st by rationalisation?

Atx\rightarrow 0
sinx\simeq x