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Math Help - Logarithmic Growth Paradox

  1. #1
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    Logarithmic Growth Paradox

    Hi - I didn't know whether this thread should go in the Chat Room, Pre-Calculus or Calculus forum, so feel free to move it to wherever appropriate.

    In studying exponential/logarithmic functions in my pre-calculus class, I realized something about logarithmic growth which lead to some sort of paradox (btw, I've taught myself a fair amount of Calculus on my own for preparation - I'm sure this question is resolved in those classes, but anyway..):

    A simple feature of the natural logarithm function, ln(x), is that as x\rightarrow\infty \Longrightarrow ln(x)\rightarrow\infty.

    Assuming the differentiation and integration formulas for transcendental functions are true without proof here, \frac{d}{dx} ln(x) = \frac{1}{x}.

    A simple feature of the reciprocal function, \frac{1}{x}, is that as x\rightarrow\infty \Longrightarrow \frac{1}{x}\rightarrow0.

    So how is it then, that ln(x)\rightarrow\infty as x\rightarrow\infty when its rate of change (on interval x > 1) modeled by its derivative, \frac{1}{x} with respect to the same argument x, goes to 0?

    In other words, how is it possible that as x\rightarrow\infty \Longrightarrow ln(x)\rightarrow\infty when as as the same time x\rightarrow\infty \Longrightarrow \frac{1}{x}\rightarrow0?

    I understand the basic principal behind logarithmic growth, that the function tends to infinity, but grows extremely slowly, and gets slower proprotional to its argument. Not unlike how the exponential function grows extremely fast, and gets faster proportional to its argument. But it's paradoxical nonetheless in regards to logarithmic growth. Exponential growth does not have this paradox, as its derivative (rate of change) goes to infinity as the function itself does ( \frac{d}{dx}e^x = e^x.).

    Not that it really matters, I just thought it was rather interesting. It relates in some sense I believe to how we define a natural logarithm as the definite integral \int\frac{1}{x} dx (bounds 1 to x - I don't know how to use Latax for definite integrals). Similarly, as x grows, the area drawn out by the curve \frac{1}{x} grows smaller and smaller as x grows larger and larger, so the logarithm function's growth is in some sense tending to 0; but how can it grow to infinity in that case as we're taught it does?
    Last edited by TaylorM0192; March 10th 2010 at 01:00 PM.
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    A thought I had to resolve it simply, is to just accept that 1/x doesn't actually reach 0, so therefore ln(x) will be increasing by something, even if that something is small. But then I thought, even though 1/x doesn't actually reach 0, the rate of change in some sense converges to 0 for ln(x), so ln(x) must converge then to some value, no?

    Just as we're taught the infinite sequence 1 + 1/n for n to infinity converges to 2?

    But we know from a number of proofs already, that ln(x) diverges to infinity, so I'm still stuck on explaining that in relation to what I posed above.
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