Given,

Lim_x->a (a*sin x - x*sin a)/(ax² - xa² )

When we put x = a in the expression, we get 0/0 form.

Now apply L. Hospital rule, we get

    Lim_x->a (a*cos x - sin a)/(2ax - a² )

= (a*cos a - sin a)/(2a*a - a² ) 

= (a*cos a - sin a)/(2a² - a² ) 

= (a*cos a - sin a)/a² 

So, Lim_x->a (a*sin x - x*sin a)/(ax² - xa² ) = (a*cos a - sin a)/a²