Question

If f(x)=2x2+3x+5 find:

Lim h->0 f(h)-f(0)/h

The answer I got is 3, but was wondering if it's right or not xD.

Answer 1

let me check

Answer 2

Okey :D thank you I'll be waiting for your reply keke

Answer 3

i got 4x+3

Answer 4

o_o doesn't it turn to 2h+3, then the h->0 makes it into 3? argh xD trying to find out how to solve these type of questions :S been awhile since I did these xD

Answer 5

but what x? u never told me the value of x

Answer 6

this is what you get when u enter f(x)
(2(x+h)^2+3(x+h)+5-(2x^2+3x+5))/h

Answer 7

then expand and distribute (2x^2+4xh+2h^2+3x+3h+5-2x^2-3x-5)/h

Answer 8

after canceling you should be left with this (2h^2+4xh+3h)/h

Answer 9

take out an H from the numerator and cancel it with the denominator
h(2h+4x+3)/h

Answer 10

then you plug in 0 for h. 2(0)+4x+3

Answer 11

4x+3

Answer 12

unless you give me a point for x i can solve it. but we dont know what x is

Answer 13

Hope this helped

Answer 14

Oh yeah thanks xD I was trying to get my head around it. But is the method that you used for derivatives? Since an answer for Limits doesn't usually end up with a result like that does it? xD

Answer 15

yes. is this question has given you a x and y point?

Answer 16

what is the question that it is asking you?