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if f(a)=2,f'(a)=1,g(a)=-1,g'(a)=-2
then lim g(x)f(a)-g(a)f(x)/x-a as x->a?
this is an indeterminate form (0/0), thus using L'Hopital's rule lim g(x)f(a)-g(a)f(x)/x-a as x→a
=(f(a)g'(a)-g(a)f'(a))/1 = (-4+1)=-3
Hi Ankit
solve it by dl hospital
diff. it and put the limit
you will get ans.
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