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Math Help - dy/dx question...

  1. #1
    Newbie DaveDammit's Avatar
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    dy/dx question...

    Hey all,

    I'm getting ready for my first semester of Calculus and I'm a big dork so I'm actually interested in the subject.

    Anyway I have a quick question: when differentiating \sqrt{x} I was taught by my trusty textbook to multiply by the conjugate using the difference quotient and so then after several calculations you end up with:

    \lim_{x \rightarrow 0}\frac{1}{\sqrt{x + h} + \sqrt{x}}

    THEN for some reason you can get rid of the "h" in \sqrt{x + h} and you're left with \frac{1}{2\sqrt{x}}.

    Now I get how to do everything just fine, I just want to know why the "h" can just go away like that.

    Thanks for the help!
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  2. #2
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    There is a mistake in your limit:
    \lim_{x \rightarrow 0}\frac{1}{\sqrt{x + h} + \sqrt{x}}

    should be:
    \lim_{h \rightarrow 0}\frac{1}{\sqrt{x + h} + \sqrt{x}}

    Do you now see why the result follows?
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  3. #3
    Newbie DaveDammit's Avatar
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    Ah yes I see I messed up the limit and should have put as "h" tends to 0 but I don't know enough about limits yet to understand why the "h" can go away.
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  4. #4
    Moo
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    Hello,
    Quote Originally Posted by DaveDammit View Post
    Ah yes I see I messed up the limit and should have put as "h" tends to 0 but I don't know enough about limits yet to understand why the "h" can go away.
    Consider this (only if there is no undetermined form, that is infinity x 0, 0/0, infinity/infinity, etc... you'll see it later)

    h goes to 0
    so x+h goes to x+0=x
    so sqrt(x+h) goes to sqrt(x) !

    that's the basic idea

    \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{\lim\limits  _{h \to 0} \sqrt{x+h}+\sqrt{x}}=\dots
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  5. #5
    Newbie DaveDammit's Avatar
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    Thumbs up

    Gotcha. Thanks to the both of you! Very helpful.
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