# 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. – Emmad Kareem Mar 11 '12 at 13:13

## 1 Answer

Both of your example are correct.

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

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 Oops!..I missed the title. The answer to your question (as in the title) is NO as explained by @Robinson, but your working out are correct. Please note also that your second example will not help to solve the first example's limit. – Tapu Mar 11 '12 at 10:52