Question

Need help with limits!! Last

Answer 1

the answer is h+6 but i dont know the steps for the alst part i know you multiply 3^2

Answer 2

does that say x approaches 0?

Answer 3

yes

Answer 4

I but that is a mistype because if that is so the answer is ((h+3)^2-3^2)/h

Answer 5

I bet they meant to write h->0

Answer 6

let's assume this

Answer 7

first do you know how to multiply out (h+3)^2?

Answer 8

oh maybe it was me but yeah

Answer 9

(h+3) * (h+3)

Answer 10

right and multiply that out
use "foil"

Answer 11

h^2 +6h+9

Answer 12

omg.

Answer 13

\[\frac{h^2+6h+9-9}{h}\]

Answer 14

now you should see 9-9 is 0

Answer 15

yeah i think i just got it

Answer 16

lol cool

Answer 17

there is other ways to look at this problem to

Answer 18

Wait nvm cus there is 3 h's how do i get ride of the last h

Answer 19

there is a common factor h on top
and a h on bottom
h/h=1

Answer 20

just like when we canceled out the x's on that one problem you had

Answer 21

Yes

Answer 22

then h+6h is left

Answer 23

\[\frac{h(h+6)}{h}=?\]

Answer 24

not exactly
we get to assume h is not 0 and say h/h is 1 here

Answer 25

h^2+6h=h(h+6)
you see this right?

Answer 26

no

Answer 27

woah i wrote thatt wrong

Answer 28

i see h*h + 6*h= h^2 +6*h

Answer 29

\[h \color{red}h+6 \color{red}h\]
both terms have an h in common

Answer 30

oh so you cross that out too?

Answer 31

you factor out the h

Answer 32

that they have in common

Answer 33

\[(h+6) \color{red}h\]