What is the name of this method that can easily find $\lim_{x \to 0^+} \frac{x^x-1}{\ln(x)+x-1}$? I can easily find $\lim_{x \to 0^+} \frac{x^x-1}{\ln(x)+x-1}$ using these steps. However, I don't completely remember the rules my professor told me about it, and I want to know what it's name is so I can look it up.
$$\lim_{x \to 0^+} \frac{x^x-1}{\ln(x)+x-1} $$
$$= \lim_{x \to 0^+} (x^x-1)\cdot\lim_{x \to 0^+}(\frac{1}{\ln(x)+x-1}) $$
$$=0 \cdot 0$$
$$=0$$
It's confusing how $\lim_{x \to 0^+}(\frac{1}{\ln(x)+x-1}) = 0$ which puzzles me since $\ln(0)$ is undefined.
 A: Referring to the puzzling part, just note that
$$\left| \frac{1}{\ln x + x -1}\right| = \frac{1}{|\ln x - (1-x)|}$$ $$\stackrel{|a-b| \geq ||a|-|b||}{\leq}\frac{1}{||\ln x| - |1-x||}\stackrel{0<x<1}{\leq} \frac{1}{||\ln x| - 1|}\stackrel{x \to 0^+}{\longrightarrow}0$$
A: Given that 
$$\lim_{x\to x_0} f(x)= f_0$$
$$\lim_{x\to x_0} g(x) = g_0$$
Where $f_0,g_0\in\mathbb{C}$ we can say that
$$\lim_{x\to x_0} \left(f(x)\cdot g(x)\right)=f_0\cdot g_0$$
A proof can be found here: http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx
A: Quickly, I'm thinking about "L'Hôpital's rule".
Basically, all you need to do is derive the top ($f(x)$) and the bottom ($g(x)$) of your equation (in term of the division), which will give you a new equation and then you will be able to replace $x$ with $0$. If it's still not a valid calculation, then the limit doesn't exists.
If you need an improved answer please tell me, but since you're looking for the rule and not the solution, I suppose it's this one.
Here's the documentation from Wikipedia : https://en.wikipedia.org/wiki/L%27Hôpital%27s_rule
A: You could have done it using Taylor series.
Consider
$$x^x=e^{x\log(x)}=1+x \log (x)+O\left(x^2\right)$$ which makes
$$ \frac{x^x-1}{\log(x)+x-1}=\frac{x \log (x)+O\left(x^2\right) }{\log(x)+x-1}=\frac{x \log (x)}{\log (x)-1}+O\left(x^2\right)\sim x$$
If the very last step is not clear, use L'Hospital rule to $\frac{x \log (x)}{\log (x)-1}$ to get $\frac{1+\log(x)}{1+\frac 1 x}=\frac{x+x\log(x)}{x+1}$
A: The rule is if $\lim_{x\to a} f(x) = k$ and $k$ is finite and $\lim_{x\to a} g(x) = m$ and $m$ is finite then $\lim_{x\to a} f(x)g(x) = km$ and $\lim_{x\to a} (f(x)+g(x))=m+k$ and if $m \ne 0$ then $\lim_{x\to a} \frac {f(x)}{g(x}) = \frac km$.
As to we $\lim_{x\to 0}\frac {1}{\ln x +x - 1} = 0$.  There is a rule that if $\lim_{x\to a} f(x) = \pm \infty$ then $\lim_{x\to a} \frac 1{f(x)} = 0$.
We can extend these rules to cases wher $\lim_{x\to a} f(x) = \infty$ but we must be careful we actually know what we are talking about or we might end up talking gibberish.
If $\lim_{x\to a}f(x) = \infty$ and $\lim_{x\to a} g(x) = k$ finite.  Then $\lim_{x\to a} (f(x) \pm g(x))=\infty$.   
If $k > 0$ then $\lim_{x\to a} f(x)g(x) = \infty$ and $\lim_{x\to a}\frac {f(x)}{g(x)} = \infty$.  If $k < 0$ then $\lim_{x\to a} f(x)g(x) = -\infty$ and $\lim_{x\to a}\frac {f(x)}{g(x)} = -\infty$. If $k=0$ then $\lim_{x\to a}\frac {g(x)}{f(x)} = 0$ but $\lim_{x\to a} f(x)g(x)$ is indeterminate.
As to youre Note:  In calculating a $\lim_{x\to 0^+} \frac 1{\ln x + x-1}$ we have $\lim_{x\to 0^+} \ln x = -\infty$ so $\lim (\ln x + x - 1)= -\infty$ and $\lim_{x\to 0^+} \frac 1{\ln x + x-1}=0$.
As to your comment that $\ln 0$ is undefined... Of course it is! That's the entire reason we are taking limits.  Notice $0^0$ is also undefined.
Note when we take limits $x \to a$, we are considering cases where $x$ is near $a$.  We are NEVER considering the the case where $x = a$.  
$f(a)$ could be undefined, could be something utterly different, or could be completely irrelevent.  We are looking at $f(x)$ when $x$ is near $a$.  So nobody cares about what $f(a)$ does.
