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jamiemm

  • 3 years ago

fine the limit if it exists: lim of x->0 [[1/(3+x)] - (1/3)]/x

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  1. 101Ryan101
    • 3 years ago
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    -(1/9) is what I got taking the derivative to the top and bottom after some algebra..

  2. 101Ryan101
    • 3 years ago
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    anyone else?

  3. satellite73
    • 3 years ago
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    that is right

  4. fiddlearound
    • 3 years ago
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    it is -1/9 http://www.wolframalpha.com/input/?i=lim+of+x-%3E0+[[1%2F%283%2Bx%29]+-+%281%2F3%29]%2Fx And click "show steps"

  5. satellite73
    • 3 years ago
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    this is the derivative of \[f(x)=\frac{1}{x}\] at x = 3 but we can do it without the derivative, just with some algebra

  6. fiddlearound
    • 3 years ago
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    hhmmmm. the link gets broken. you'll need to copy/paste it .

  7. satellite73
    • 3 years ago
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    \[\frac{1}{3+x}-\frac{1}{3}=\frac{3-(3+x)}{x(3+x)}\] \[=\frac{-x}{3(3+x)}\] divide by x get \[\frac{-1}{3(3+x)}\] replace x by 0 get \[-\frac{1}{9}\]

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