anonymous
  • anonymous
what is the limit of (3+h)^-1-3-1/h as h approaches 0?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
is it maybe \[\frac{(3+h)^{-1}-\frac{1}{3}}{h}\]
anonymous
  • anonymous
no my fault its (3+h)^-1+3^-1/h
anonymous
  • anonymous
ok well that is fine, because it is the same thing, as \(3^{-1}=\frac{1}{3}\)

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anonymous
  • anonymous
in fact the first step is to get rid of the negative exponent, and do some algebra
anonymous
  • anonymous
|dw:1378435055117:dw|
anonymous
  • anonymous
no
anonymous
  • anonymous
awesome
anonymous
  • anonymous
lets go slow and also lets forget the \(h\) in the denominator for a second
anonymous
  • anonymous
\[(3+h)^{-1}=\frac{1}{3+h}\] and \[3^{-1}=\frac{1}{3}\] so your numerator is \[\frac{1}{3+h}-\frac{1}{3}\]
anonymous
  • anonymous
first job is to subtract (we will come back to the \(h\) in the denominator later)
anonymous
  • anonymous
ok
anonymous
  • anonymous
get a common denominator
anonymous
  • anonymous
yeah and don't multuply out, leave in factored form you get \[\frac{3-(3+h)}{3(3+h)}\]
anonymous
  • anonymous
as always, everything in the top without an \(h\) goes, leaving you with \[\frac{-h}{3(3+h)}\]
anonymous
  • anonymous
now recall that there was an \(h\) in the bottom so we put that back
anonymous
  • anonymous
or just cancel, either way \[\frac{-h}{3(3+h)h}=\frac{-1}{3(3+h)}\]
anonymous
  • anonymous
let me know if and where i lost you, because there is only one more step, namely take the limit by replacing \(h\) by \(0\)
anonymous
  • anonymous
i got it thank you... have a tough time with algebra rules that havent been learned in years
anonymous
  • anonymous
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 \(-\frac{1}{9}\)
anonymous
  • anonymous
haha hopefully i get that good

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