Property of Natural Log- Inequality equation

[chetanladha]

Hi.

I just saw on wikipedia that natural logarithm has such a property:
[x/(1+x)] < ln (1 + x) < x

(http://en.wikipedia.org/wiki/Natural_logarithm)

Can anyone pls tell me how to prove this?

Proving [x/(1+x)] and ln (1 + x) less than 'x' is easy.. But how abt [x/(1+x)] < ln (1 + x) ??

Thanks in advance..

 

[dextercioby]

It's wrong. It's smaller or equal.

Consider the function

$$f:(-1,+\infty)\longrightarrow \mathbb{R}$$

$$f(x) = \frac{x}{x+1} - \ln (1+x)$$

Plot it.

 

[BloodyFrozen]

Following what dextercioby said, the following inequality does not always hold true.

[x/(1+x)] < ln (1 + x) < x

For example, let x = 0. We the have:

0/(1+0) < ln(1+0) < 0
0 < 0 < 0

This makes no sense.

 

[chetanladha]

Thanks for response.

Yes its smaller or equal (Sorry couldn't rite it..)..

Is plotting the function f(x) = [x/(1+x)] - ln (1 + x) best way 2 prove it??

 

[dextercioby]

You needn't plot it, just make the standard analysis of 1st and 2nd derivatives of f to be able to draw the desired conclusion.

 

[chetanladha]

Thanks a lot..!