Question

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

Answer 1

is it maybe
\[\frac{(3+h)^{-1}-\frac{1}{3}}{h}\]

Answer 2

no my fault its (3+h)^-1+3^-1/h

Answer 3

ok well that is fine, because it is the same thing, as \(3^{-1}=\frac{1}{3}\)

Answer 4

in fact the first step is to get rid of the negative exponent, and do some algebra

Answer 5

no

Answer 6

awesome

Answer 7

lets go slow and also lets forget the \(h\) in the denominator for a second

Answer 8

\[(3+h)^{-1}=\frac{1}{3+h}\] and
\[3^{-1}=\frac{1}{3}\] so your numerator is
\[\frac{1}{3+h}-\frac{1}{3}\]

Answer 9

first job is to subtract (we will come back to the \(h\) in the denominator later)

Answer 10

ok

Answer 11

get a common denominator

Answer 12

yeah and don't multuply out, leave in factored form
you get
\[\frac{3-(3+h)}{3(3+h)}\]

Answer 13

as always, everything in the top without an \(h\) goes, leaving you with
\[\frac{-h}{3(3+h)}\]

Answer 14

now recall that there was an \(h\) in the bottom so we put that back

Answer 15

or just cancel, either way
\[\frac{-h}{3(3+h)h}=\frac{-1}{3(3+h)}\]

Answer 16

let me know if and where i lost you, because there is only one more step, namely take the limit by replacing \(h\) by \(0\)

Answer 17

i got it thank you... have a tough time with algebra rules that havent been learned in years

Answer 18

yeah don't fret too much
this is the first or second week of calculus
by the third week you will say
"this is the derivative of \(\frac{1}{x}\) which is \(-\frac{1}{x^2}\) if \(x=3\) so the answer has to be \(-\frac{1}{9}\)

Answer 19

haha hopefully i get that good