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Thread: Limits

  1. #1
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    Limits

    Given that $\displaystyle \lim_{x \to 0}~\frac{e^x-1}{x}=1$

    Evaluate:

    a) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+x)}}{x}$

    b) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+5x)}}{x}$

    I'm not sure how to do these ones?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nzmathman View Post
    Given that $\displaystyle \lim_{x \to 0}~\frac{e^x-1}{x}=1$

    Evaluate:

    a) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+x)}}{x}$

    b) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+5x)}}{x}$

    I'm not sure how to do these ones?
    Well, you can apply L'Hopital's rule. it doesn't use what we were given though...
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  3. #3
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    I forgot to include one thing...these questions followed on from a question proving that $\displaystyle \lim_{x \to 0}~\frac{f(x)}{x}= \lim_{x \to 0}~\frac{x}{f^{-1}(x)}$ provided that f is an invertible function.

    By the way we are instructed not to use that rule as it is not in the syllabus...
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  4. #4
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    Quote Originally Posted by nzmathman View Post
    Given that $\displaystyle \lim_{x \to 0}~\frac{e^x-1}{x}=1$

    Evaluate:

    a) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+x)}}{x}$

    b) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+5x)}}{x}$

    I'm not sure how to do these ones?
    or this method

    $\displaystyle \left. a \right){\text{ }}\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \left\{ \begin{gathered}x = \frac{1}{t} \Leftrightarrow \frac{1}{x} = t \hfill \\x \to 0,{\text{ }}t \to \infty \hfill \\ \end{gathered} \right\} =$

    $\displaystyle = \mathop {\lim }\limits_{t \to \infty } \ln {\left( {1 + \frac{1}{t}} \right)^t} = \ln e = 1.$

    $\displaystyle \left. b \right){\text{ }}\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 5x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \ln {\left( {1 + 5x} \right)^{\frac{1}{x}}} = \left\{ \begin{gathered}5x = \frac{1}{t} \Leftrightarrow \frac{1}{x} = 5t \hfill \\ x \to 0,{\text{ }}t \to \infty \hfill \\ \end{gathered} \right\} =$

    $\displaystyle = \mathop {\lim }\limits_{t \to \infty } \ln {\left( {1 + \frac{1}{t}} \right)^{5t}} = 5 \cdot \mathop {\lim }\limits_{t \to \infty } \ln {\left( {1 + \frac{1}{t}} \right)^t} = 5\ln e = 5.$
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nzmathman View Post
    I forgot to include one thing...these questions followed on from a question proving that $\displaystyle \lim_{x \to 0}~\frac{f(x)}{x}= \lim_{x \to 0}~\frac{x}{f^{-1}(x)}$ provided that f is an invertible function.

    By the way we are instructed not to use that rule as it is not in the syllabus...
    ok, now we're getting somewhere.

    note that the inverse of $\displaystyle y = \ln (1 + x)$ is $\displaystyle y = e^x - 1$

    so the first limit is equal to $\displaystyle \lim_{x \to 0} \frac x{e^x - 1}$, and the rest is easy


    now for the second one: what's the inverse of $\displaystyle \ln (1 + 5x)$ ?
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  6. #6
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    I finally got it now!

    I get the inverse as $\displaystyle \frac{e^x-1}{5}$ and it follows from there that the limit I need to find is equivalent to $\displaystyle 5\cdot\frac{1}{\lim_{x \to 0} \frac{e^x - 1}{x}} = 5$, and that answer is confirmed by my calculator
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nzmathman View Post
    I finally got it now!

    I get the inverse as $\displaystyle \frac{e^x-1}{5}$ and it follows from there that the limit I need to find is equivalent to $\displaystyle 5\cdot\frac{1}{\lim_{x \to 0} \frac{e^x - 1}{x}} = 5$
    yes

    and that answer is confirmed by my calculator
    and DeMath's post
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