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Math Help - l'H˘pital's rule

  1. #1
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    l'H˘pital's rule

    Hi,how can i solve by l'H˘pital's rule
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  2. #2
    Super Member girdav's Avatar
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    Do you really have to solve it by l'H˘pital rule ? You can write \sqrt{9x^2+1}  =3x\sqrt{1+\frac 1{9x^2}} and \sqrt{4x^2+1}  =2x\sqrt{1+\frac 1{4x^2}}.
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  3. #3
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    Quote Originally Posted by girdav View Post
    Do you really have to solve it by l'H˘pital rule ? You can write \sqrt{9x^2+1}  =3x\sqrt{1+\frac 1{9x^2}} and \sqrt{4x^2+1}  =2x\sqrt{1+\frac 1{4x^2}}.
    thanks , yes please by l'H˘pital rule
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  4. #4
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    The derivative of (9x^2+ 1)^{1/2} is
    \frac{1}{2}(9x^2+ 1)^{-1/2}(18x)= \frac{9x}{\sqrt{9x^2+ 1}}

    The derivative of (4x^2+ 1)^{1/2} is
    \frac{1}{2}(4x^2+ 1}^{-1/2}(8x)= \frac{4x}{\sqrt{4x^2+ 1}}

    So the ratio of the derivatives is
    \frac{9x}{\sqrt{9x^2+ 1}}\frac{\sqrt{4x^2+ x}}{4x}= \frac{9}{4}\frac{\sqrt{4x^2+ 1}}{\sqrt{9x^2+ 1}}

    But that is just 9/4 times the reciprocal of the original fraction. If, by L'Hopital's rule, we must have the limit of the original fraction equal to the limit of that, we must have, calling the limit of the original fraction L, L= (9/4)(1/L). Can you solve that for L? Of course, technically, you should also prove that L'Hopital's rule actually works here- that the limit exists.
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  5. #5
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    Quote Originally Posted by BAHADEEN View Post
    thanks , yes please by l'H˘pital rule
    Why would anyone use l'H˘pital's rule on that limit?
    What a waste of time, let alone discouraging understanding.
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