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Thread: evaluating a limit with ln

  1. #1
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    evaluating a limit with ln

    Hi

    So the limit is

    lim x ( ln (x+5) - ln x)
    x-> infinity

    so everything goes to infinity as x goes to infinity but I don't think I can use l'Hospital's rule b/c it is not in quotient form.

    Where do I go from here to evaluate the limit?

    Thanks

    Calculus beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    Hi

    So the limit is

    lim x ( ln (x+5) - ln x)
    x-> infinity

    so everything goes to infinity as x goes to infinity but I don't think I can use l'Hospital's rule b/c it is not in quotient form.

    Where do I go from here to evaluate the limit?

    Thanks

    Calculus beginner
    $\displaystyle \lim_{x \to \infty}\left(x (\ln(x+5)-\ln(x)\right) = \lim_{x \to \infty} \ln\left( \left(\dfrac{x+5}{x}\right)^x\right) $

    Since $\displaystyle \lim_{x \to \infty} \left( \left(1+\dfrac{5}{x}\right)^x\right) = e^5$

    your limit exists and has the value 5.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by calcbeg View Post
    Hi

    So the limit is

    lim x ( ln (x+5) - ln x)
    x-> infinity

    so everything goes to infinity as x goes to infinity but I don't think I can use l'Hospital's rule b/c it is not in quotient form.

    Where do I go from here to evaluate the limit?

    Thanks

    Calculus beginner
    $\displaystyle \mathop {\lim }\limits_{x \to \infty } x\left( {\ln \left( {x + 5} \right) - \ln x} \right) = \mathop {\lim }\limits_{x \to \infty } x\ln \left( {1 + \frac{5}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } \ln {\left( {1 + \frac{5}{x}} \right)^x} =$

    $\displaystyle = \left\{ \begin{gathered}
    x = 5t, \hfill \\
    x \to \infty , \hfill \\
    t \to \infty \hfill \\
    \end{gathered} \right\} = \mathop {\lim }\limits_{t \to \infty } \ln {\left( {1 + \frac{1}
    {t}} \right)^{5t}} = \ln {e^5} = 5.$
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  4. #4
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    Hello, calcbeg!

    Find the limit: .$\displaystyle \lim_{x\to\infty}x
    \bigg[\ln(x+5) - \ln x\bigg] $

    We have: .$\displaystyle \lim_{x\to\infty}x\bigg[\ln(x+5) - \ln(x)\bigg] \;=\;\lim_{x\to\infty}\bigg[\frac{\ln(x+5) - \ln(x)}{\dfrac{1}{x}}\bigg] \quad\rightarrow\quad \frac{0}{0}$

    Apply L'Hopital: .$\displaystyle \lim_{x\to\infty}\left[\frac{\frac{1}{x+5} - \frac{1}{x}}{-\frac{1}{x^2}}\right] \;=\;\lim_{x\to\infty}\frac{5x}{x+5}$

    Divide top and bottom by $\displaystyle x\!:\;\;\lim_{x\to\infty}\frac{\frac{5x}{x}}{\frac {x}{x} + \frac{5}{x}} \;=\;\lim_{x\to\infty}\frac{5}{1+\frac{5}{x}} \;=\;\frac{5}{1+0} \;=\;5$

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