# Proving!

• Nov 29th 2012, 08:46 PM
Tutu
Proving!
Hi I have two questions here, which I do not know how to prove.

1.) Prove that (lnx)/x ≤ 1/e for all x>0.

I know to find the maximum value of (lnx)/x, I got x=e. But then what do I do? I subbed x=e back into (lnx)/x and I got 1/e.. 1/e ≤ 1/e doesn't make sense..

2.) Prove that lnx ≤ x-1 for all x>0.
Do I find the maximum value for lnx? I can't really get the maximum value for I'll get 1 = 0.

• Nov 29th 2012, 08:52 PM
Prove It
Re: Proving!
Question 2 just requires you to know that \displaystyle \begin{align*} \lim_{x \to 0^{+}} \ln{(x)} = -\infty \end{align*} while \displaystyle \begin{align*} \lim_{x \to 0^{+}} (x - 1) = -1 \end{align*}, which means \displaystyle \begin{align*} \ln{(x)} \end{align*} starts out less than \displaystyle \begin{align*} x - 1 \end{align*}, and also that \displaystyle \begin{align*} \ln{(x)} \end{align*} grows slower than \displaystyle \begin{align*} x-1 \end{align*}.
• Nov 30th 2012, 03:02 AM
emakarov
Re: Proving!
Quote:

Originally Posted by Tutu
1/e ≤ 1/e doesn't make sense.

Why does is not make sense?
• Nov 30th 2012, 04:20 AM
coolge
Re: Proving!
If the maximum value of the function is 1/e then obviously for all x the function value is < 1/e.
You cannot find the maximum value for ln x. It is a monotonically increasing function that asymptotically goes to \infinity.
• Nov 30th 2012, 06:14 AM
Plato
Re: Proving!
Quote:

Originally Posted by Tutu
2.) Prove that lnx ≤ x-1 for all x>0.

There is a famous inequality.
Use the mean value theorem to prove that.
If $0 then $\frac{1}{b}\le\frac{\ln(b)-\ln(a)}{b-a}\le\frac{1}{a}$.