I am having a problem with the calculation of the following limit.
I need to find: $\lim_{x \to 0^{+}} (\ln \frac{1}{x})^x$
Thank you in advance
I am having a problem with the calculation of the following limit.
I need to find: $\lim_{x \to 0^{+}} (\ln \frac{1}{x})^x$
Thank you in advance
Take $f(x)=\ln^x(1/x)$ so as @Eugene suggested take $\ln$ of both sides, so $\ln(f(x))=x\ln(\ln(1/x))$ or $\ln(f(x))=\frac{\ln(\ln(1/x))}{1/x}$ Now take $1/x=t$ so when $x$ tends to $0^+$; $t$ tends to $+\infty$. By L'Hospital's Rule you have $\ln(f(x))\to 0$ so $f(x)\to 1$.
It may be easier to look at large numbers. So let $w=1/x$. We study $(\ln w)^{1/w}$. Take the logarithm of this. We get $\dfrac{\ln(\ln w)}{w}$.
By L'Hospital's Rule, or otherwise, this approaches $0$ as $w\to\infty$. So our limit is $1$.
For the L'Hospital's Rule argument, differentiating top and bottom yields $\dfrac{1}{w\ln w}$, which clearly has limit $0$.