If `f(x)` is differentiable `EE f'(1) = 4, f'(2) = 6` then `lim_(h rarr 0) (f(3h+2+h^3)-f(2))/(f(2h - 2h^2 + 1)-f(1))=`

If f(x) is differentiable EE f'(1)=4,f'(2)=6 then lim_(h rarr0)(f(3h+2+h^(3))-f(2))/(f(2h-2h^(2)+1)-f(1))=

f(x) is differential function. Given f^(')(1) =4, f^(')(2) - 6 , then lim_(h rarr 0) (f(2+2h+h^(2))-f(2))/(f(1+h-h^(2))-f(1))

If f'(3)=2, then lim_(h rarr0)(f(3+h^(2))-f(3-h^(2)))/(2h^(2)) is

If f'(2)=1 , then lim_(h to 0)(f(2+h)-f(2-h))/(2h) is equal to

If f(x) is a differentiable function of x then lim_(h rarr0)(f(x+3h)-f(x-2h))/(h)=

If f'(3)=2 , then lim_(h->0)(f(3+h^2)-f(3-h^2))/(2h^2) is

Let f be differentiable at x=0 and f'(0)=1 Then lim_(h rarr0)(f(h)-f(-2h))/(h)=

If f(x) =1/x , then show that lim_(hrarr0)(f(2+h)-f(2))/h=-1/4

Given that f'(2)=6 and f'(1)=4 then {:("Lim"),(hrarr0):}(f(2h+2+h^(2))-f(2))/(f(h-h^(2)+1)-f(1)) is :