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I have posted a link there to this topic so the OP can see my work.A not so simple problem, or is it?

x^ln x/ (lnx)^x ; when x approaches infinity

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- Thread starter
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- #1

I have posted a link there to this topic so the OP can see my work.A not so simple problem, or is it?

x^ln x/ (lnx)^x ; when x approaches infinity

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We are given to compute:

\(\displaystyle L=\lim_{x\to\infty}\frac{x^{\ln(x)}}{\ln^x(x)}\)

Observing that we have the indeterminate form \(\displaystyle \frac{\infty}{\infty}\), we may apply L'Hôpital's Rule. Thus, we need to compute the derivatives of the numerator and denominator.

Let's begin with the numerator, and let:

\(\displaystyle y=x^{\ln(x)}\)

Taking the natural log of both sides, we obtain:

\(\displaystyle \ln(y)=\ln^2(x)\)

Implicitly differentiating with respect to $x$, we have:

\(\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{2}{x}\ln(x)\)

\(\displaystyle \frac{dy}{dx}=\frac{2y}{x}\ln(x)\)

\(\displaystyle \frac{dy}{dx}=\frac{2x^{\ln(x)}}{x}\ln(x)=2x^{\ln(x)-1}\ln(x)\)

Now, for the denominator, let:

\(\displaystyle y=\ln^x(x)\)

Taking the natural log of both sides, we have:

\(\displaystyle \ln(y)=x\ln(\ln(x))\)

Implicitly differentiating with respect to $x$, we have:

\(\displaystyle \frac{1}{y}\frac{dy}{dx}=x\frac{1}{\ln(x)}\frac{1}{x}+\ln(\ln(x))=\ln(\ln(x))+\frac{1}{\ln(x)}\)

\(\displaystyle \frac{dy}{dx}=y\left(\ln(\ln(x))+\frac{1}{\ln(x)} \right)\)

\(\displaystyle \frac{dy}{dx}=\ln^x(x)\left(\ln(\ln(x))+\frac{1}{\ln(x)} \right)\)

And so we may now state:

\(\displaystyle L=\lim_{x\to\infty}\frac{2x^{\ln(x)-1}\ln(x)}{\ln^x(x) \left(\ln(\ln(x))+\frac{1}{\ln(x)} \right)}= 2L\lim_{x\to\infty}\frac{\ln(x)}{x\left(\ln(\ln(x))+\frac{1}{\ln(x)} \right)}\)

At this point, we may observe that the denominator dominates the numerator, and so we may conclude:

\(\displaystyle L=0\)

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\(\displaystyle L=\lim_{x\to\infty}e^{\ln(x)(\ln(x)-x)}\)

Hence:

\(\displaystyle \ln(L)=\lim_{x\to\infty}\left(\ln(x)(\ln(x)-x) \right)=\lim_{x\to\infty}\left(\frac{\frac{\ln(x)}{x}-1}{\frac{1}{x\ln(x)}} \right)=-\infty\)

and so:

\(\displaystyle L=0\)