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.
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
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 $k &gt; 0$ then $\lim_{x\to a} f(x)g(x) = \infty$ and $\lim_{x\to a}\frac {f(x)}{g(x)} = \infty$.  If $k &lt; 0$ 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.