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ChrisV

lim x>0- ((x+deltaX)^2+x+deltaX-(x^2+x))/deltaX

  • 2 years ago
  • 2 years ago

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  1. ChrisV
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    \[\lim_{x \rightarrow 0-} ((x+\Delta x)^2 +x + \Delta x-(x^2+x)/\Delta x\]

    • 2 years ago
  2. Rogue
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    \[\lim_{x \rightarrow 0-} \frac {(x+\Delta x)^2 +x + \Delta x-(x^2+x)}{\Delta x}\]Your gonna have to simplify that out, although it gets annoying...

    • 2 years ago
  3. Rogue
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    \[\lim_{\Delta x \rightarrow 0} \frac {(x+\Delta x)^2 +x + \Delta x-(x^2+x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x - x^2 - x}{\Delta x}\]

    • 2 years ago
  4. ChrisV
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    yea i did that my problem is after that

    • 2 years ago
  5. ChrisV
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    i cancel out the x^2 and X

    • 2 years ago
  6. Rogue
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    \[\lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x + \Delta x - x^2 - x}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { 2x \Delta x + \Delta x + \Delta^2 x}{\Delta x} = \lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x\]

    • 2 years ago
  7. ChrisV
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    2xdeltax+deltax^2+deltaX/deltax

    • 2 years ago
  8. Rogue
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    \[\lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x = 2x + 1 + 0 = 2x + 1\]

    • 2 years ago
  9. Rogue
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    Yeah, divide the delta x's out to simplifiy. Then just evaluate the limit by plugging in 0 for delta x.

    • 2 years ago
  10. ChrisV
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    ok so i factor the deltax out

    • 2 years ago
  11. ChrisV
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    deltax(2x+deltax+1)/1

    • 2 years ago
  12. ChrisV
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    oops i mean over delta x not 1

    • 2 years ago
  13. ChrisV
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    so i end up with 2x+deltax+1

    • 2 years ago
  14. ChrisV
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    set delta x to 0

    • 2 years ago
  15. ChrisV
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    gives me 2x+1

    • 2 years ago
  16. ChrisV
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    however the book gives me an answer of -1/x^2

    • 2 years ago
  17. ChrisV
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    so im doing something wrong

    • 2 years ago
  18. ChrisV
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    so im confused beyond belief atm

    • 2 years ago
  19. Rogue
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    -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.

    • 2 years ago
  20. ChrisV
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    the instruction is to find the limit if it exist, if it does not explain why

    • 2 years ago
  21. ChrisV
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    so how do they get a limit of 1/x^2 out of 2x+1?

    • 2 years ago
  22. ChrisV
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    well the book does not show me the work in the back, just the answer

    • 2 years ago
  23. ChrisV
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    guess this is one to ask the instructor :)

    • 2 years ago
  24. Rogue
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    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.

    • 2 years ago
  25. ChrisV
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    k thanks

    • 2 years ago
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