Question

computing limits,from example 2,how did he get the -6h

Answer 1

what is the example?

Answer 2

@mine where is example 2??

What is your question??

Answer 3

having any trouble posting ?

Answer 4

If yes, then contact hartnn..

Ha ha ha...

Answer 5

oh yeah, message me ;)

Answer 6

Simply write the question here..

You can write it in words too...

Answer 7

Lim2[-3+h]2-18
h

Answer 8

._.

Answer 9

That is something we can see..

Answer 10

thats good,
i think its
\(\huge \lim \limits_{x->2}\frac{[-3+h]^2-18}{h}\)
right ?

Answer 11

Otherwise I am wondering what could be the example 2??

How it looks like!!

Ha ha..

Answer 12

Is that right @mine ??

Answer 13

sorry, h->2

Answer 14

That sounds good..

Answer 15

no, more like
\(\huge \lim \limits_{h->0}\frac{2[-3+h]^2-18}{h}\)

Answer 16

now that makes sense as a Question ^

Answer 17

if h->2 then we can directly substitute

Answer 18

but if h->0...

Answer 19

h should be 0 there...

Answer 20

Where?

Answer 21

then numerator, we get
[2(h^2-6h+9) -18 ]/h
are you talking about this '-6h' ?

Answer 22

I means limit should be from h tends to 0 @ParthKohli

Answer 23

anyways, continuing...
(2h^2-12h)/h = 2h-12 = -12 , when h->0

Answer 24

yeah,-6

Answer 25

\[(a-b)^2 = a^2 + b^2 - 2ab\]

Do you know this??

Answer 26

yeah

Answer 27

Then apply this formula there..

Answer 28

with a=h and b=3

Answer 29

\[(-3+h)^2 \implies (h-3)^2 = ??\]

Answer 30

k,tanks

Answer 31

Can you go further???

If any problem arises, then message @hartnn

Answer 32

hope all your doubts are clear...any more doubts ?

Answer 33

yeah