Question

Find the limit.
\[\lim_{h \rightarrow 0}\frac{ \frac{ 1 }{ (3+h)^2 }-\frac{ 1 }{ 9 } }{ h }\]

Answer 1

HI!!

Answer 2

how far have you gotten in calculus?

Answer 3

MMM... WHAT WAS YOUR FUNCTION TO BEGIN WITH ?

Answer 4

Well, right now we're at extreme values, but we having a mid-term next week. So we're reviewing.

Answer 5

k, now my question (lol)

Answer 6

if you have gotten to derivatives, you recognize this as the derivative of \(f(x)=\frac{1}{x^2}\) at \(x=3\)

Answer 7

take the derivative of \(f(x)=\frac{1}{x^2}\) then find \(f'(3)\)
if you have not gotten to derivatives you have more work to do

Answer 8

Can we not do this with derivatives, I have to review for limits. So the long way to solve this...

Answer 9

@SolomonZelman I'm not sure what you mean. This is question, there is no other functions for this. That's the original...

Answer 10

oh, that was the original question ?

Answer 11

\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{\frac{1}{(3+h)^2}-\frac{1}{9}}{h}}\)

find the common denominator and add the fractions on the top.

Answer 12

Yes, I did that

Answer 13

and what do you get after doing this ?

Answer 14

\[\lim_{h \rightarrow 0}\ \frac{ \frac{ -6h-h^2 }{ 9(9+6h+h^2) } }{ h }\]

Answer 15

\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{\frac{1}{(3+h)^2}-\frac{1}{9}}{h}}\)


\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{\frac{9}{9(3+h)^2}-\frac{(h+3)^2}{9(h+3)^2}}{h}}\)

\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{\frac{9-(h+3)^2}{9(3+h)^2}}{h}}\)

\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{9-(h+3)^2}{9h(3+h)^2}}\)

\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{3^2-(h+3)^2}{9h(3+h)^2}}\)

\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(3-h-3)(3+h+3)}{9h(3+h)^2}}\)

Answer 16

like this I am using a difference of squares (in my last step)

Answer 17

(no need to expand the (3+h)^2 as you did, even though at times it appears to be an intuitive thing to do)

Answer 18

if you still have questions, ask...

Answer 19

okay... and then what? Sub in the 0, but it would be indeterminate...

Answer 20

3-h-3=?

Answer 21

0?

Answer 22

3-h-3
3-3-h
0-h
-h



so 3-h-3=-h

Answer 23

\[\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(3-h-3)(3+h+3)}{9h(3+h)^2}} \\ \large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(-h)(3+h+3)}{9h(3+h)^2}} \\ \]

Answer 24

you should see something that cancels

Answer 25

OH! I see! the h's reduce and it would be -6/81 --> -2/27

Answer 26

yah

Answer 27

yes, it is correct.
freckles tnx for continuing for me.

Answer 28

Thank you! @SolomonZelman and @freckles

Answer 29

yw