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Math Help - A limit as x approaches a.

  1. #1
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    A limit as x approaches a.

    \lim_{x \to a} \dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}

    There aren't any steps for this on Wolfram so..

    I have a question. I know that this limit is to find the derivative at a certain point, whereas x \to 0 is for any point. I must have forgotten how to evaluate this but.. How exactly do I do this? Can I use direct substitution? I'm not looking for the answer or for you to show me how you would do it, just where I start. Thanks!
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  2. #2
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    Since this goes to \displaystyle \frac{0}{0} you can use L'Hospital's Rule.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Since this goes to \displaystyle \frac{0}{0} you can use L'Hospital's Rule.

    Yes but since the derivatives of those constant a's are zero isn't that a problem?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by qleeq View Post
    \lim_{x \to a} \dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}

    There aren't any steps for this on Wolfram so..

    I have a question. I know that this limit is to find the derivative at a certain point, whereas x \to 0 is for any point. I must have forgotten how to evaluate this but.. How exactly do I do this? Can I use direct substitution? I'm not looking for the answer or for you to show me how you would do it, just where I start. Thanks!
    I don't see how this limit could represent the derivative of some function. So to evaluate this, you could try to use L'H˘pital's rule, since \lim\limits_{x\to a}\dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}\rightarrow\dfrac{0}{0}.

    Edit: Prove It beat me to it.
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  5. #5
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    They are not constant functions, they are coefficients of the functions of \displaystyle x, which is fine...
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    Quote Originally Posted by Chris L T521 View Post
    I don't see how this limit could represent the derivative of some function. So to evaluate this, you could try to use L'H˘pital's rule, since \lim\limits_{x\to a}\dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}\rightarrow\dfrac{0}{0}.

    Edit: Prove It beat me to it.
    Sorry, this is only my first year of calculus and we haven't had a formal lecture on L'Hopital's rule yet, so I'm only just now learning how it works.
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  7. #7
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    Quote Originally Posted by qleeq View Post
    Sorry, this is only my first year of calculus and we haven't had a formal lecture on L'Hopital's rule yet, so I'm only just now learning how it works.
    That's fine...

    Basically \displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} as long as your original function gives \displaystyle \frac{0}{0} or \displaystyle \frac{\infty}{\infty} from direct substitution.

    I can show you a proof of this if you like.
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  8. #8
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    I have a feeling I'm over-complicating this..

    \lim_{x \to a} \dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}

    \lim_{x \to a} \dfrac{\dfrac{1}{2}(2a^3x-x^4)^{-1/2} - \dfrac{1}{3}a(a^2x)^{-2/3}}{a - \frac{1}{4}(ax^3)^{-3/4}}

    \lim_{x \to a} \dfrac{\dfrac{1}{2}(2a^3x-x^4)^{-1/2}\*(2a^3 - 4x^3) - \dfrac{1}{3}a(a^2x)^{-2/3}\*a^2}{a - \dfrac{1}{4}(ax^3)^{-3/4}\*(3x^2a)}
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  9. #9
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    You haven't done the differentiating properly... You need to use the Chain Rule for each term...
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  10. #10
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    Quote Originally Posted by Prove It View Post
    You haven't done the differentiating properly... You need to use the Chain Rule for each term...

    Lame. Where exactly have I messed up?
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  11. #11
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    Each term you've differentiated is incorrect...
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  12. #12
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    Figured just as much. I think the thing that's throwing me off is the 'a'.. how do I deal with it?
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  13. #13
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    \displaystyle a gets treated as a constant.
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  14. #14
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    No, it is NOT the "a" that is throwing you off. What is throwing you haven't used the chain rule as Prove It said.

    You have the derivative of \sqrt{2a^3x- x^4}= \left(2a^3x- x^4\right)^{1/2} as \frac{1}{2}\left(2a^3x- x^5\right)^{-1/2}. That's wrong because, by the chain rule, you must multiply that by the derivative of 2a^x- x^4 with respect to x.
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  15. #15
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    Quote Originally Posted by HallsofIvy View Post
    No, it is NOT the "a" that is throwing you off. What is throwing you haven't used the chain rule as Prove It said.

    You have the derivative of \sqrt{2a^3x- x^4}= \left(2a^3x- x^4\right)^{1/2} as \frac{1}{2}\left(2a^3x- x^5\right)^{-1/2}. That's wrong because, by the chain rule, you must multiply that by the derivative of 2a^x- x^4 with respect to x.

    After some trying..

    \dfrac{\dfrac{a^3-2x^3}{(2a^3x-x^4)^{1/2}}-{\dfrac{a^3}{(3a^2x)^{2/3}}}}{-\dfrac{3ax^2}{4(ax^3)^{3/4}}}

    Is this closer?
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