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Thread: [SOLVED] limit of the natural logarithm

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] limit of the natural logarithm

    How do you prove that $\displaystyle \lim_{x \to +\infty}ln(x) = +\infty$? Starting from the fact that $\displaystyle ln(x)=\int_1^x \frac{dt}{t}$, so without using anything related to $\displaystyle e^x$.
    And also that $\displaystyle \lim_{x \to 0}ln(x) = -\infty$.
    I just don't understand what my teacher did... Meaningless to me.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by arbolis View Post
    How do you prove that $\displaystyle \lim_{x \to +\infty}ln(x) = +\infty$? Starting from the fact that $\displaystyle ln(x)=\int_1^x \frac{dt}{t}$, so without using anything related to $\displaystyle e^x$.
    And also that $\displaystyle \lim_{x \to 0}ln(x) = -\infty$.
    I just don't understand what my teacher did... Meaningless to me.
    Since $\displaystyle \frac{1}{t}$ is decreasing its min is at the right endpoint of any interval.

    $\displaystyle
    \int_{i}^{i+1}\frac{1}{t}dt > 1\cdot \left( \frac{1}{i+1} \right) \forall \\\ i \in \mathbb{N}
    $

    Summing both sides of the inequality gives

    $\displaystyle \sum_{i=1}^{n}\int_{i}^{i+1}\frac{1}{t}dt > \sum_{i=1}^{n}\left( \frac{1}{i+1} \right) $

    $\displaystyle \int_{1}^{n}\frac{1}{t}dt > \sum_{i=1}^{n}\frac{1}{i+1}=\infty$ as $\displaystyle n \to \infty$
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks! I understand what you did.
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    to see this
    Attached Thumbnails Attached Thumbnails [SOLVED] limit of the natural logarithm-m.jpg  
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  5. #5
    MHF Contributor arbolis's Avatar
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    Thanks invite moon! My professor did something similar to this, now I understand him.
    In your explanation, you should also show that $\displaystyle ln(x)$ is increasing, and that $\displaystyle ln3>1$.
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    Sorry ,I thought that ln(x) is increasing and that ln3>1,are well known things.
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  7. #7
    MHF Contributor arbolis's Avatar
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    Yes there are. For instance, $\displaystyle ln(e)=1$, it can be shown that $\displaystyle e<3$. The derivative of $\displaystyle ln(x)$ is $\displaystyle \frac{1}{x}$, always positive in the dominion of the ln function, so ln is increasing.
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  8. #8
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    Quote Originally Posted by arbolis View Post
    How do you prove that $\displaystyle \lim_{x \to +\infty}ln(x) = +\infty$? Starting from the fact that $\displaystyle ln(x)=\int_1^x \frac{dt}{t}$, so without using anything related to $\displaystyle e^x$.
    And also that $\displaystyle \lim_{x \to 0}ln(x) = -\infty$.
    I just don't understand what my teacher did... Meaningless to me.
    First note that natural logarithm is an incresing function (1) since $\displaystyle (\ln(x))'=\frac{1}{x}>0$ for all $\displaystyle x>0$

    Now: $\displaystyle \ln(x)=\int_1^x\frac{du}{u}=\int_a^{a\cdot{x}}\fra c{du}{u}=\ln(a\cdot{x})-\ln(a)$

    Thus the Natual Logarithm has the following property, for $\displaystyle a,b>0$:$\displaystyle \ln(a)+\ln(b)=\ln(a\cdot{b})$

    Thus: $\displaystyle n\cdot{\ln(2)}=\underbrace{\ln(2)+...+\ln(2)}_n=\l n(2^n)$ for any natural number $\displaystyle n$ (2)

    Suppose we want to prove that $\displaystyle \lim_{x \to +\infty}\ln(x) = +\infty
    $ by definition.

    We have to prove that, for all
    $\displaystyle h>0$ there is a number $\displaystyle \delta_{h}>0$ such that if $\displaystyle x>\delta_{h}$ we have $\displaystyle \ln(x)>h$ (3)

    Setting $\displaystyle \delta_{h}=2^{\frac{h}{\ln(2)}}$ by (1) and (2) we know that the condition (3) holds.

    Just see that $\displaystyle \ln(2^n)=n\cdot{\ln(2)}>h$ iff $\displaystyle n>\frac{h}{\ln(2)}$ and since the function is increasing if $\displaystyle x>2^n$ then $\displaystyle \ln(x)>\ln(2^n)>h$

    So by definition it follows that $\displaystyle \lim_{x \to +\infty}\ln(x) = +\infty
    $ (4)


    For the second part set
    $\displaystyle u=\frac{1}{x}$ and use the property shown in the second line, so that $\displaystyle \ln(x)=-\ln(u)$ and since $\displaystyle x\to{0^+}$ we have $\displaystyle u \to +\infty$ and the assertion follows by (4)
    Last edited by PaulRS; Jul 10th 2008 at 12:54 PM.
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  9. #9
    MHF Contributor arbolis's Avatar
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    Nicely done PaulRS. Thank you.
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    Quote Originally Posted by TheEmptySet View Post
    $\displaystyle \sum_{i=1}^{n}\frac{1}{i+1}=\infty$ as $\displaystyle n \to \infty$
    The problem with this approach is that the typical proof of the divergence of the harmonic series uses the integral test which uses $\displaystyle \lim_{x\to \infty}\ln x = \infty$. So it is kind of circular. But not really because there are some proofs which show divergence of harmonic series without integral tests.

    Quote Originally Posted by PaulRS View Post
    Setting $\displaystyle \delta_{h}=2^{\frac{h}{\ln(2)}}$ by (1) and (2) we know that the condition (3) holds.
    Good job. But there is just a minor problem here. What does $\displaystyle 2^{h/\ln 2}$ mean? You are assuming the definition of exponents for arbitrary real powers. However, the generalized definition needs further properties of $\displaystyle \ln$ which you are trying to avoid in the first place.
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  11. #11
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    Quote Originally Posted by arbolis View Post
    How do you prove that $\displaystyle \lim_{x \to +\infty}ln(x) = +\infty$? Starting from the fact that $\displaystyle ln(x)=\int_1^x \frac{dt}{t}$, so without using anything related to $\displaystyle e^x$.
    i'll prove it using the definition of limit. let $\displaystyle M > 0.$ put $\displaystyle N=\lfloor{M \rfloor} + 1,$ and choose $\displaystyle K=2^{2N}.$ if $\displaystyle x > K,$ then:

    $\displaystyle \ln x = \int_1^x \frac{dt}{t} > \int_1^K \frac{dt}{t} = \sum_{j=0}^{2N - 1} \int_{2^j} ^{2^{j+1}}\frac{dt}{t} > \sum_{j=0}^{2N-1} \int_{2^j}^{2^{j+1}} \frac{dt}{2^{j+1}}=\sum_{j=0}^{2N-1} \frac{1}{2}=N > M. \ \ \ \square$

    And also that $\displaystyle \lim_{x \to 0}ln(x) = -\infty$.
    I just don't understand what my teacher did... Meaningless to me.
    let $\displaystyle \frac{1}{t}=u.$ then $\displaystyle \ln x = \int_1^x \frac{dt}{t} = -\int_1^{\frac{1}{x}} \frac{du}{u} = - \ln \left(\frac{1}{x}\right).$ now apply the first part of your problem.
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    Here is a general theorem. Let $\displaystyle f$ be a function (defined on $\displaystyle \mathbb{R}^+$) such that (i)$\displaystyle f$ is increasing (ii)$\displaystyle f$ is not bounded from above. Then $\displaystyle \lim_{x\to \infty}f(x) = \infty$. Here is the proof. We prove this directly by definition. Let $\displaystyle A>0$. Then since $\displaystyle f$ is not bounded from above there is $\displaystyle B>0$ such that $\displaystyle f(B)>A$. Then if $\displaystyle x>B$ it means $\displaystyle f(x) > f(B) > A$. Proof complete. Now we can apply this to $\displaystyle \ln$ function. Since $\displaystyle (\ln x)' = \tfrac{1}{x}>0$ this function is increasing this satisfies (i). Now we need to satisfy (ii) i.e. the function is unbounded this takes a little bit more work. Notice that (when $\displaystyle x>1$) $\displaystyle \smallint_1^x t^{-1}dt \geq \smallint_1^x t^{-3/2} dt = 2 -2x^{-1/2}$ and $\displaystyle 2 - 2(9)^{-1/2} = \tfrac{4}{3} > 1$. Thus, $\displaystyle \ln (9) > 1$. Since $\displaystyle 0= \ln 1<1$ and $\displaystyle \ln 9>1$ it means there is a number, call it $\displaystyle e$, such that $\displaystyle \ln e = 1$ (by property of continous functions). Then we see as PaulRS showed that $\displaystyle \ln e^n = n$ for all positive integers $\displaystyle n$. But since we can make $\displaystyle n$ are large as we please it means $\displaystyle \ln$ is an unbounded function. The rest follows.
    Last edited by ThePerfectHacker; Jul 11th 2008 at 09:44 AM.
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