Answers

2016-02-13T16:40:46+05:30
Let l=lim(x->a) (xsina - asinx) / (x-a) 
this is of form 0/0...............substituting x=a 

therefore, we can apply L' Hospital's Rule; so differentiating numerator and denominator respectively. 

=>l= lim(x->a) (sina - acosx)/1 
=>l=sina - acosa 
which is the required limit.
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The Brainliest Answer!
2016-02-13T16:44:48+05:30
  
Limit x -> a   x sina - asinx / x - a
 Limit x -> a   xsina - asinx + asina - asina / x - a

Limit x -> a  ( x - a ) sina  / x - a      - a ( sinx - sina ) / x - a

Limit x -> a      sina     -a Limit x -> a  ( sinx - sina ) / x - a                         ( ∴sinc - sind = 2 cos(c+d) / 2  sin(c-d ) /2 )

Limit x -> a   sina   -a Limit x -> a   2cos( x+a) /2 sin( x - a) /2  / x - a

sina         -2acos( a+a) /2  Limit x -> a    sin (x -a )/ 2 / x - a

sina     -2acos( 2a ) /2      Limit x - a /2 -> a - a/2   sin (x -a )/ 2 / x - a /2  x 1/2

sina  -2acosa     1 x 1/2

sina - acosa

  solved ..
hope its help u
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