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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.2012-03-11
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    No, they applied [L'Hôpital's rule](http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule).2012-03-11
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    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.$2012-03-11
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    Please fix the title of your question. Thanks.2012-03-11

1 Answers 1

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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.2012-03-11