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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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    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