Why is $\infty^0$ indeterminate? In a recent test question I was required to us L'Hopital's rule to evaluate:

$$\lim_{x\to 0^+} x\ln{(e^{2x}-1)}$$

I assumed that anything multiplied by 0 would give an answer of 0. This turns out not to be the case. Is there a simple explanation as to why infinity multiplied by 0 is not 0?
Any number, when multiplied by 0, gives 0.  However, infinity is not a real number.  When we write something like $\infty \cdot 0$, this doesn't directly mean anything; rather, it's shorthand for a certain type of limit, where the first part approaches infinity.

Now, zero times anything approaching $\infty$ will still give a limit of zero.  However, that's not what the shorthand $\infty \cdot 0$ means.  It means something approaching infinity multiplied by something approaching zero.  And this doesn't have to be zero at all.  

For a simple example, as $x \rightarrow \infty$, $x^2$ certainly approaches infinity.  And $\frac{1}{x^2}$ certainly approaches zero.  But $x^2 \cdot \frac{1}{x^2} = 1$, so when we multiply the two together we get something approaching 1 (because it is constantly 1).

In essence, solving these problems boils down to figuring out whether the part approaching infinity grows fast enough to "cancel out" the part approaching zero, or if it's the other way around, or if they grow/shrink at rates that perfectly match each other (as is the case with $x^2$ and $\frac{1}{x^2}$).
Here's very simple case: $\lim\limits_{x\to 0+} x\cdot\frac{6}{x}$.  Clearly $x$ goes to $0$.

But $x\cdot\frac{6}{x} = 6$ whenever $x\neq0$. So $\lim\limits_{x\to 0+} x\cdot\frac{6}{x} = \lim\limits_{x\to0+} 6 = 6$.  This limit is not $0$.