- #1

- 9

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter karanmohan
- Start date

- #1

- 9

- 0

- #2

- 15,393

- 686

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.

- #3

- 9

- 0

- #4

- 15,393

- 686

- #5

- 1,552

- 15

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

- 354

- 2

- #7

- 12,121

- 160

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

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

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

Share: