Question

what is the limit of (3-h)^-1 - 3^-1 / h as h approaches 0

Answer 1

\[\lim \frac{ (3-h)^{-1} - 3^{-1}}{ h }\]

Answer 2

as h approaches 0

Answer 3

is th answer: indeterminate?

Answer 4

No, you just need to manipulate it a bit:
\[\begin{align*}\lim_{h\to0}\frac{\dfrac{1}{3-h}-\dfrac{1}{3}}{h}&=\lim_{h\to0}\frac{\dfrac{3}{3(3-h)}-\dfrac{3-h}{3(3-h)}}{h}\\
&=\lim_{h\to0}\frac{1}{h}\left(\frac{3-(3-h)}{3(3-h)}\right)\\
\end{align*}\]

Answer 5

so the limit is 1

Answer 6

That's not what I'm getting, no.

Answer 7

you would canel out the (3-h)

Answer 8

*cancel

Answer 9

No, you wouldn't, that's not how dividing works:\[\lim_{h\to0}\frac{1}{h}\left(\frac{3-(3-h)}{3(3-h)}\right)=\lim_{h\to0}\frac{1}{h}\left(\frac{h}{3(3-h)}\right)=\lim_{h\to0}\frac{1}{3(3-h)}\]

Answer 10

kk. that makes sense. Im stupid

Answer 11

Don't worry about it, just takes some practice :)

Answer 12

isnt there supposed to be a (-) in front of the h in the numerator?

Answer 13

oh wait, is it 3-3+h?

Answer 14

It's \(3-(3-h)=3-3+h=h\).

Answer 15

so the answer is 1/9

Answer 16

Yes, that's right

Answer 17

thank you so much.