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Math Help - how is this happening?

  1. #1
    Junior Member
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    how is this happening?



    <br />
\frac{\ln{\frac{x+3}{x+1}}}{1/x} = \frac{2x^2}{(x + 1)(x + 3)}<br />

    This is related to trying to get lim->infinity so technically it is calculus
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  2. #2
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    Hello, TYTY!

    \lim_{x\to\infty} \frac{\ln{\frac{x+3}{x+1}}}{\frac{1}{x}} \:=\: \lim_{x\to\infty} \frac{2x^2}{(x + 1)(x + 3)}
    The limit goes to \tfrac{0}{0} . . . so we can apply L'Hopital


    We have: . \frac{\ln (x+3) - \ln(x+1)}{x^{-1}}

    Apply L'Hopital: . \frac{\frac{1}{x+3}-\frac{1}{x+1}} {-x^{-2}} \;=\;\frac{\frac{-2}{(x+1)(x+3)}}{-\frac{1}{x^2}} \;=\;\frac{2x^2}{(x+1)(x+3)}

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  3. #3
    Junior Member
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    Quote Originally Posted by Soroban View Post
    Hello, TYTY!

    The limit goes to \tfrac{0}{0} . . . so we can apply L'Hopital


    We have: . \frac{\ln (x+3) - \ln(x+1)}{x^{-1}}

    Apply L'Hopital: . \frac{\frac{1}{x+3}-\frac{1}{x+1}} {-x^{-2}} \;=\;\frac{\frac{-2}{(x+1)(x+3)}}{-\frac{1}{x^2}} \;=\;\frac{2x^2}{(x+1)(x+3)}

    Good lord I think my brain just exploded.

    you rock soroban
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