anonymous
  • anonymous
lim x>0- ((x+deltaX)^2+x+deltaX-(x^2+x))/deltaX
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\lim_{x \rightarrow 0-} ((x+\Delta x)^2 +x + \Delta x-(x^2+x)/\Delta x\]
Rogue
  • Rogue
\[\lim_{x \rightarrow 0-} \frac {(x+\Delta x)^2 +x + \Delta x-(x^2+x)}{\Delta x}\]Your gonna have to simplify that out, although it gets annoying...
Rogue
  • Rogue
\[\lim_{\Delta x \rightarrow 0} \frac {(x+\Delta x)^2 +x + \Delta x-(x^2+x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x - x^2 - x}{\Delta x}\]

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anonymous
  • anonymous
yea i did that my problem is after that
anonymous
  • anonymous
i cancel out the x^2 and X
Rogue
  • Rogue
\[\lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x + \Delta x - x^2 - x}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { 2x \Delta x + \Delta x + \Delta^2 x}{\Delta x} = \lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x\]
anonymous
  • anonymous
2xdeltax+deltax^2+deltaX/deltax
Rogue
  • Rogue
\[\lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x = 2x + 1 + 0 = 2x + 1\]
Rogue
  • Rogue
Yeah, divide the delta x's out to simplifiy. Then just evaluate the limit by plugging in 0 for delta x.
anonymous
  • anonymous
ok so i factor the deltax out
anonymous
  • anonymous
deltax(2x+deltax+1)/1
anonymous
  • anonymous
oops i mean over delta x not 1
anonymous
  • anonymous
so i end up with 2x+deltax+1
anonymous
  • anonymous
set delta x to 0
anonymous
  • anonymous
gives me 2x+1
anonymous
  • anonymous
however the book gives me an answer of -1/x^2
anonymous
  • anonymous
so im doing something wrong
anonymous
  • anonymous
so im confused beyond belief atm
Rogue
  • Rogue
-1/x^2 is the answer to another question, probably the derivative of 1/x. The answer to the question you posted here is 2x + 1. There might be an error in the book, or you read it wrong.
anonymous
  • anonymous
the instruction is to find the limit if it exist, if it does not explain why
anonymous
  • anonymous
so how do they get a limit of 1/x^2 out of 2x+1?
anonymous
  • anonymous
well the book does not show me the work in the back, just the answer
anonymous
  • anonymous
guess this is one to ask the instructor :)
Rogue
  • Rogue
No, for your question, 2x + 1 is the correct answer. That's all there is to it. You used the limit formula to find the derivative of x^2 + x.\[\frac {dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac {f(x + \Delta x) - f(x)}{\Delta x}\] If you used the same formula for the function 1/x, you would get -1/x^2. You are correct, the textbook is wrong. Don't worry too much about it.
anonymous
  • anonymous
k thanks

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