# Is $\lim_{x\to \infty} \frac{\ln{x}}{x} =\lim_{x\to \infty} \frac{\frac{1}{x}}{1}$?

In my notes its given

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

Is that correct? How do I get that?

I think another example is also related

$$x^{\frac{1}{x}} = e^{\frac{\ln{x}}{x}}$$

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The first example is correct and is L'Hospital's rule. – Jeff Mar 11 '12 at 10:27
No, they applied L'Hôpital's rule. – Raskolnikov Mar 11 '12 at 10:28
It isn't true that that equality holds for all $x.$ But the conditions to apply L'Hopital's rule hold, so this allows the differentation of the numerator and denominator, leaving the limit of the ratio unchanged. So while the individual terms are different, both ratios have the same limit as $x \to \infty.$ – Geoff Robinson Mar 11 '12 at 10:38
Please fix the title of your question. Thanks. – NoChance Mar 11 '12 at 13:13

Please note that $\ln x\to\infty$ as $x\to\infty$. Therefore, $$\lim_{x\to\infty}=\frac{\ln x}{x}\equiv\frac{\infty}{\infty},$$ an indeterminate form. Hence by L' Hospital rule, $$\lim_{x\to\infty}=\frac{\ln x}{x}=\frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x)}=\frac{\frac{1}{x}}{1}$$
Your second example: we have $x^{\frac{1}{x}} = e^{\ln x^\frac{1}{x}}=e^{\frac{\ln x}{x}}$.