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Xcessive's question at Yahoo! Answers regarding a limit
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Hello Xcessive,
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:
\(\displaystyle L=0\)
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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Another approach would be to write the limit as:
\(\displaystyle L=\lim_{x\to\infty}e^{\ln(x)(\ln(x)-x)}\)
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:
\(\displaystyle L=0\)
\(\displaystyle L=\lim_{x\to\infty}e^{\ln(x)(\ln(x)-x)}\)
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:
\(\displaystyle L=0\)