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## Answers

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.

The Brainliest Answer!

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