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Math Help - [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 \lim_{x \to +\infty}ln(x) = +\infty? Starting from the fact that ln(x)=\int_1^x \frac{dt}{t}, so without using anything related to e^x.
    And also that \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 \lim_{x \to +\infty}ln(x) = +\infty? Starting from the fact that ln(x)=\int_1^x \frac{dt}{t}, so without using anything related to e^x.
    And also that \lim_{x \to 0}ln(x) = -\infty.
    I just don't understand what my teacher did... Meaningless to me.
    Since \frac{1}{t} is decreasing its min is at the right endpoint of any interval.

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

    Summing both sides of the inequality gives

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

    \int_{1}^{n}\frac{1}{t}dt > \sum_{i=1}^{n}\frac{1}{i+1}=\infty as 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 ln(x) is increasing, and that 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, ln(e)=1, it can be shown that e<3. The derivative of ln(x) is \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 \lim_{x \to +\infty}ln(x) = +\infty? Starting from the fact that ln(x)=\int_1^x \frac{dt}{t}, so without using anything related to e^x.
    And also that \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 (\ln(x))'=\frac{1}{x}>0 for all x>0

    Now: \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 a,b>0: \ln(a)+\ln(b)=\ln(a\cdot{b})

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

    Suppose we want to prove that \lim_{x \to +\infty}\ln(x) = +\infty<br />
by definition.

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

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

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

    So by definition it follows that \lim_{x \to +\infty}\ln(x) = +\infty<br />
(4)


    For the second part set
    u=\frac{1}{x} and use the property shown in the second line, so that \ln(x)=-\ln(u) and since x\to{0^+} we have u \to +\infty and the assertion follows by (4)
    Last edited by PaulRS; July 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
    \sum_{i=1}^{n}\frac{1}{i+1}=\infty as 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 \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 \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 2^{h/\ln 2} mean? You are assuming the definition of exponents for arbitrary real powers. However, the generalized definition needs further properties of \ln which you are trying to avoid in the first place.
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    Quote Originally Posted by arbolis View Post
    How do you prove that \lim_{x \to +\infty}ln(x) = +\infty? Starting from the fact that ln(x)=\int_1^x \frac{dt}{t}, so without using anything related to e^x.
    i'll prove it using the definition of limit. let M > 0. put N=\lfloor{M \rfloor} + 1, and choose K=2^{2N}. if x > K, then:

    \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 \lim_{x \to 0}ln(x) = -\infty.
    I just don't understand what my teacher did... Meaningless to me.
    let \frac{1}{t}=u. then \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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  12. #12
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    Here is a general theorem. Let f be a function (defined on \mathbb{R}^+) such that (i) f is increasing (ii) f is not bounded from above. Then \lim_{x\to \infty}f(x) = \infty. Here is the proof. We prove this directly by definition. Let A>0. Then since f is not bounded from above there is B>0 such that f(B)>A. Then if x>B it means f(x) > f(B) > A. Proof complete. Now we can apply this to \ln function. Since (\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 x>1) \smallint_1^x t^{-1}dt \geq \smallint_1^x t^{-3/2} dt = 2 -2x^{-1/2} and 2 - 2(9)^{-1/2} = \tfrac{4}{3} > 1. Thus, \ln (9) > 1. Since 0= \ln 1<1 and \ln 9>1 it means there is a number, call it e, such that \ln e = 1 (by property of continous functions). Then we see as PaulRS showed that \ln e^n = n for all positive integers n. But since we can make n are large as we please it means \ln is an unbounded function. The rest follows.
    Last edited by ThePerfectHacker; July 11th 2008 at 09:44 AM.
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