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Use the principle that  when  lim x->0, if  u - >0 and v ->0 ,
     then Lim x->0 u / v =   lim  x ->0 ,  u' / v'       (limit of the ratio of derivatives)

So we use this principle twice in the following steps.

 \lim_{x \to 0} \frac{(Sin\ x- x) }{x\ Sin\ x}= \lim_{x \to 0} \frac{(Cos\ x- 1) }{(x\ Cos\ x+Sinx)}\\\\=\lim_{x \to 0} \frac{-Sin\ x }{(-x\ Sin\ x + Cos\ x+ Cos\ x)}\\\\= \frac{- 0 }{(-0+1+1)} = 0\\

  • Brainly User
see,in order to calulate the limit differentiate the given expression since limit is tending to 0
you willget differentiation of 1/x as -1/x^2 ans 1/sinx is cosecx and so differentiation of cosecx is -cosecx.cotx and now put x=0 and you willget the answer as 0