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Math Help - indeterminant form please help

  1. #1
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    indeterminant form please help

    ok so this is what it says to do

    evaluate lim (sinx)/ (x^(1/2)) as x goes to 0+

    ok so i've used LHopital's rule but what i got was dne. i think i did something wrong but i dont know what... here's what i did

    lim (cosx)/((1/2) x ^(-1/2)) as x--> 0+ = 1/0
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  2. #2
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    Hello, runner07!

    Remember this theorem? . . \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1


    Evaluate: . \lim_{x\to0^+}\frac{\sin x}{x^{\frac{1}{2}}}

    We have: . \frac{\sin x}{\sqrt{x}}

    Multiply by \frac{\sqrt{x}}{\sqrt{x}}\!:\;\;\frac{\sqrt{x}}{\s  qrt{x}}\cdot\frac{\sin x}{\sqrt{x}} \;=\;\sqrt{x}\cdot\frac{\sin x}{x}

    Then: . \lim_{x\to0^+}\left[\sqrt{x}\cdot\frac{\sin x}{x}\right] \;=\;\sqrt{0}\cdot1 \;=\;0

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  3. #3
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    OH! i forgot about that theorem. but that makes more sense. thank you!!
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  4. #4
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    wat

    wat is the answer???????

    evaluate lim (sinx)/ (x^(1/2)) as x goes to 0+

    ok so i've used LHopital's rule but what i got was dne. i think i did something wrong but i dont know what... here's what i did

    lim (cosx)/((1/2) x ^(-1/2)) as x--> 0+ = 1/0[/quote]
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  5. #5
    GAMMA Mathematics
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    Quote Originally Posted by kritica View Post
    wat is the answer???????

    evaluate lim (sinx)/ (x^(1/2)) as x goes to 0+

    ok so i've used LHopital's rule but what i got was dne. i think i did something wrong but i dont know what... here's what i did

    lim (cosx)/((1/2) x ^(-1/2)) as x--> 0+ = 1/0
    When using L'Hopital's rule, you will always be dividing by zero since the power of one half will never disappear. Follow the eloquent solution that Soroban provided.

    EDIT: Actually, never mind, I had an algebraic brain fart! Look at topsquark's post below!
    Last edited by colby2152; May 22nd 2008 at 11:29 AM.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kritica View Post
    wat is the answer???????

    evaluate lim (sinx)/ (x^(1/2)) as x goes to 0+

    ok so i've used LHopital's rule but what i got was dne. i think i did something wrong but i dont know what... here's what i did

    lim (cosx)/((1/2) x ^(-1/2)) as x--> 0+ = 1/0
    It can be done by L'Hopital's rule. But watch that negative exponent in the denominator.

    \lim_{x \to 0^+} \frac{sin(x)}{\sqrt{x}}

    = \lim_{x \to 0^+} \frac{cos(x)}{\frac{1}{2\sqrt{x}}}
    ala L'Hopital

    = \lim_{x \to 0^+} 2 \sqrt{x} ~ cos(x) = 0

    -Dan
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Also it is known for very small values of x \sin(x)\sim{x}

    which means you can replace the \sin(x) with x in this case givign

    \lim_{x\to{0}}\frac{x}{\sqrt{x}}=\lim_{x\to{0}}\sq  rt{x}=0
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