Question

Given f(x)=2x^2-1, evaluate lim of h->0 (f(1+h) -f(1))/(h)

Answer 1

Answer is 4.

Answer 2

Thanks! but how do you get to it?

Answer 3

the question is basically the first principle method for finding derivatives with x = 1...so u just take the derivative of the given equation and substitute x as 1.
First principle-
f ' (x) = lim h-->0 [f(x+h)-f(h)]/h

Answer 4

f(1+h) -f(1) = 2(1+h)^2-1 -(2(1)^2-1)
= 2(1+h)^2-1 -1 = 2[(1+h)^2-1]
=2(h^2+2h+1-1) = 2(h^2+2h)
now (f(1+h) -f(1))/h = 2(h+2)
therefore limh->0 (f(1+h) -f(1))/h =2()+2)=4
And 4 is urs ans