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karanmohan

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- Thread starter karanmohan
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karanmohan

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However, there is a lot one can say about log(x+1) for large x. Write x+1=x(1+1/x). Then log(x+1) = log(x(1+1/x)) = log(x) + log(1+1/x) ~ log(x) + 1/x.

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karanmohan

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- #5

lugita15

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This was something that always puzzled me when I was taking calculus. As x increases, the graph of ln(x) "flattens out," but it flattens out so incredibly slowly that it still manages to cross every single horizontal line y=C. It's just such a counter-intuitive phenomenon. Can anyone shed any light into this mystery?

The same intuition that makes people think that ln(x) should have a horizontal asymptote would also make them think that e^x has a vertical asymptote somewhere. How is it that a function which becomes infinitely steep as x goes to infinity manages to still cross every vertical line x=C?

- #6

Anonymous217

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I guess I don't see what the mystery is, the part that I boldfaced actually sums it up very well. By the way, the square root function does this too -- though ln(This was something that always puzzled me when I was taking calculus.As x increases, the graph of ln(x) "flattens out," but it flattens out so incredibly slowly that it still manages to cross every single horizontal line y=C.It's just such a counter-intuitive phenomenon. Can anyone shed any light into this mystery?

But if you want a real mind-blower, look at a plot of ln(ln(

- #8

Hurkyl

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1 + (1/2) + (1/3) + (1/4) + (1/5) + ...

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