MHB Xcessive's question at Yahoo Answers regarding a limit

[MarkFL]

Here is the question:

A not so simple problem, or is it?

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

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Xcessive,

We are given to compute:

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

Observing that we have the indeterminate form $$\frac{\infty}{\infty}$$, 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:

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

Taking the natural log of both sides, we obtain:

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

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

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

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

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

Now, for the denominator, let:

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

Taking the natural log of both sides, we have:

$$\ln(y)=x\ln(\ln(x))$$

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

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

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

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

And so we may now state:

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

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

$$L=0$$

 

[MarkFL]

Another approach would be to write the limit as:

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

Hence:

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

and so:

$$L=0$$