Question #ba1d3

1 Answer
Sep 10, 2017

e) 8x-3

Explanation:

When you see (x+h) in the first principles formula, wherever you'd normally put x put (x+h)

lim_(h->0)(4(x+h)^2-3(x+h)+5-(4x^2-3x+5))/h substituting

lim_(h->0)(4(x^2+h^2+2xh)-3x-3h+5-4x^2+3x-5)/h expanding

lim_(h->0)(4x^2+4h^2+8xh-3x-3h+5-4x^2+3x-5)/h

quite a lot of these terms will cancel:

lim_(h->0)(4h^2+8xh-3h)/h

= lim_(h->0)4h+8x-3

now look at the limit - as h goes to zero, what happens?

4(0)+8x-3 = 8x-3