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

1. ## 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

2. $\lim_{x \to \infty}\frac{\ln{x}}{x}$.

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

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

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

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

$= 0$.

3. oh DUHHHH because infinite/infinite DOES NOT equal 1; it is indeterminate!!!

4. 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

5. 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 $x \ge e$

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

so

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

Thus,

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

which gives

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

A non L'Hopital way of doing it.

6. 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.

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# lim x goes.to infinity of ln[x/x 1]

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