# why does the limit of ln(x)/x approach 0???

• Aug 28th 2010, 07:14 AM
RedSwiss
why does the limit of ln(x)/x approach 0???
the limit of ln(x) is infinite, so why does dividing it by x make it zero? If anything it should make it 1 because infinite/infinite = 1
• Aug 28th 2010, 07:27 AM
Prove It
$\displaystyle \lim_{x \to \infty}\frac{\ln{x}}{x}$.

Since this goes to $\displaystyle \frac{\infty}{\infty}$ you can use L'Hospital's Rule.

$\displaystyle \lim_{x \to \infty}\frac{\ln{x}}{x} = \lim_{x \to \infty}\frac{\frac{d}{dx}(\ln{x})}{\frac{d}{dx}(x) }$

$\displaystyle = \lim_{x \to \infty}\frac{\frac{1}{x}}{1}$

$\displaystyle = \lim_{x \to \infty}\frac{1}{x}$

$\displaystyle = 0$.
• Aug 28th 2010, 07:29 AM
RedSwiss
oh DUHHHH because infinite/infinite DOES NOT equal 1; it is indeterminate!!!
• Aug 28th 2010, 07:31 AM
skeeter
Quote:

Originally Posted by RedSwiss
the limit of ln(x) is infinite, so why does dividing it by x make it zero? If anything it should make it 1 because infinite/infinite = 1

• Aug 28th 2010, 07:45 AM
Jester
Quote:

Originally Posted by RedSwiss
the limit of ln(x) is infinite, so why does dividing it by x make it zero? If anything it should make it 1 because infinite/infinite = 1

Even though both numerator and denominator are growing without bound, the real question is which is growing faster.

Also note that for $\displaystyle x \ge e$

$\displaystyle 1 \le \ln x \le \sqrt{x}$

so

$\displaystyle \dfrac{1}{x} \le \dfrac{\ln x}{x} \le \dfrac{1}{\sqrt{x}}$,

Thus,

$\displaystyle \displaystyle \lim_{x \to \infty} \dfrac{1}{x} \le \lim_{x \to \infty} \dfrac{\ln x}{x} \le \lim_{x \to \infty} \dfrac{1}{\sqrt{x}}$

which gives

$\displaystyle \displaystyle \lim_{x \to \infty} \dfrac{\ln x}{x}} = 0.$

A non L'Hopital way of doing it.
• Aug 28th 2010, 09:38 AM
HallsofIvy
Quote:

Originally Posted by RedSwiss
oh DUHHHH because infinite/infinite DOES NOT equal 1; it is indeterminate!!!

Well, I'm glad you said that. It would have sounded harsh if I had said it!

Note that ln(x) and x both go to infinity as x goes to infinity. But x goes to infinity faster than ln(x) does.