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I have to calculate the following limit:

$$\lim_{x\to \infty} 2x^{1/\ln x}$$

So I tried to start:

$$\lim_{x\to \infty} 2x^{1/ \ln x} = \infty^0 $$ From here on I noticed that I have to use de l'hopital rule. but don't really know how and I need help.

If the math representation doesn't clear, so the exerice is:

2x^(1/ln x) as x $\to \infty$

thanks in advance.

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Try calculating the limit of the logarithm of your expression. – ronash Jun 3 '13 at 11:50
Are you sure you've read the question correctly? $x^{\frac 1{\ln x}} = e$ so the limit is trivial. – Tim Jun 3 '13 at 11:51
up vote 1 down vote accepted

This is almost trivial and doesn't require l'Hospital:

$$2x^{\frac1{\log x}}=2e^{\frac1{\log x}\log x}$$

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

and thus

$$\lim_{x\to\infty}2x^{\frac1{\log x}}=2e$$

Perhaps your function is mistyped?

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Thank you very much! – Billie Jun 3 '13 at 12:25

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