L'Hospital's rule problem $\lim_{x\to 0^+}(x^{x}-1)\ln(x)$ $$\lim_{x\to 0^+}(x^{x}-1)\ln(x)$$ I need to solve this by L´Hopital´s rule: this is an indetermination of the type $0 \cdot \infty$:
$$\lim_{x\to 0^+}(x^{x}-1)\ln(x)=\lim_{x\to 0^+}{(x^{x}-1)\over {1\over \ln(x)}} $$
and this is an indetermination of the type $0/0$, so by L'Hospital's rule:
$$\lim_{x\to 0^+}{(x^{x}-1)\over {1\over \ln(x)}}
=\lim_{x\to 0^+}{(x^{x}(1+\ln(x))\over {-1\over x\ln^{2}(x)}}$$
and this is an indetermination of the type $\infty/\infty$. but if I keep on using L'Hopital's rule, the limit will just get bigger so how can I do to solve this?
 A: We start as in the post. We need to know that we are dealing with an indeterminate form. For that we need to know that $\lim_{x\to 0^+} x^x=1$. That can be done by rewriting $x^x$ as $e^{x\ln x}$, and finding the limit of $\frac{\ln x}{1/x}$ using L'Hospital's  Rule.
Thus, precisely as in the post, we find that we want 
$$\lim_{x\to 0^+} (x^x) (-x)(\ln^2 x+\ln^3 x).$$
Now it's nearly over. The term $x^x$ behaves nicely, we can forget about it. So we need 
$$\lim_{x\to 0^+}-x(\ln^2 x+\ln^3 x).$$
Let's deal with the $x\ln^3 x$ part. We use, what else, L'Hospital's Rule, and a familiar trick.  
Rewrite $x\ln^3 x$ as $\frac{\ln^3 x}{1/x}$. By using L'Hospital's Rule, we find $$\lim_{x\to 0^+}\frac{\ln^3 x}{1/x}=\lim_{x\to 0^+} -3x\ln^2 x.$$ Now do it again. We want $\lim_{x\to 0^+} 6x\ln x$. And we know what this is. 
Remark: There are savings possible. For instance, rewrite $x\ln^3 x$ as
$\left(\dfrac{\ln x}{x^{-1/3}}\right)^3$. L/Hospital's Rule, applied once, shows that $\dfrac{\ln x}{x^{-1/3}}$ has limit $0$.  
A: Hint:
$$\lim_{x\to 0^+}(x^{x}-1)\ln(x)\\
\implies \text{Let $x=\dfrac{1}{t}$, giving }\lim_{t\to +\infty}\left(\left(\dfrac{1}{t}\right)^{1/t}-1\right)\ln\left(\dfrac{1}{t}\right)\\
\implies \lim_{t\to +\infty}\left(\dfrac{1-t^{1/t}}{t^{1/t}}\right)\ln\left(\dfrac{1}{t}\right)\\
\implies \lim_{t\to +\infty}\left(\dfrac{(1-t^{1/t})\ln\left(\dfrac{1}{t}\right)}{t^{1/t}}\right)\\
\implies \lim_{t\to +\infty}\dfrac{\left(x^{1/x}-1\right)x+x^{1/x}\ln(x)}{-x^{\frac{1}{x}-2}}$$
You can show that this goes to zero as $t$ goes to infinity.
A: If there is a logarithm, always choose to differentiate it, otherwise it doesn't simplify at all. Do it the other way:
$$\lim_{x\to 0^+}\frac{\ln x}{1/(x^x-1)}=$$
$$=\lim_{x\to 0^+}\frac{1/x}{-1/(x^x-1)^2 \frac{d}{dx}(e^{x\ln x}-1)}=$$
$$=\lim_{x\to 0^+}\frac{-(x^x-1)^2}{x e^{x\ln x}(\ln x+1)}=$$
$$=\lim_{x\to 0^+}\frac{-(x^x-1)^2}{x^x (x\ln x+x)}=$$
$$=\lim_{x\to 0^+}\frac{1}{x^x}\lim_{x\to 0^+}\frac{-(x^x-1)^2}{x\ln x+x}=$$
$$=\lim_{x\to 0^+}\frac{-2(x^x-1)x^x(\ln x +1)}{\ln x + 2}=$$
$$=-2\lim_{x\to 0^+}(x^x-1)x^x\lim_{x\to 0^+}\frac{\ln x +1}{\ln x + 2}=-2\times 0\times 1=0$$
I'm sure it can be done more elegantly.
