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?
 A: "Infinity times zero" or "zero times infinity" is a "battle of two giants". Zero is so small that it makes everyone vanish, but infinite is so huge that it makes everyone infinite after multiplication. In particular, infinity is the same thing as "1 over 0", so "zero times infinity" is the same thing as "zero over zero", which is an indeterminate form.
Your title says something else than "infinity times zero". It says "infinity to the zeroth power". It is also an indefinite form because
$$\infty^0 = \exp(0\log \infty) $$
but $\log\infty=\infty$, so the argument of the exponential is the indeterminate form "zero times infinity" discussed at the beginning. By the way, in many cases, you are right that the argument will be zero because $\log\infty$ is a "smaller" infinity than the normal infinity, and the zero will "beat it". But there is no universal rule: the result will depend on the functions.
Zero is also the winner in your particular homework problem.
$$\exp(2x)-1 = 2x+O(x^2)$$
as $x\to 0$, so $\log$ of the argument above is $\log(2x)$ which goes to $-\infty$ but in a slower way than $x$ goes to zero, so the product of $x$ and the logarithm goes to zero as $x\to 0$.
A: It's indeterminate because it can be anything you like! Consider these three limits:
$$\lim_{x\to\infty} x \frac{1}{x} = \lim_{x\to\infty} 1 = 1$$
$$\lim_{x\to\infty} x^2 \frac{1}{x} = \lim_{x\to\infty} x = \infty$$
$$\lim_{x\to\infty} x \frac{1}{x^2} = \lim_{x\to\infty} \frac{1}{x} = 0$$
All of them are superficially of the form $\infty$ times $0$, but the results are very different! About the only thing you can say with certainty is that the result won't be negative if the factors are positive (a 'positive indeterminate' if you like).
By simplifying expression such as these to statements about $\infty$ and $0$, you throw away information about the rates at which the quantities involved go to infinity or zero; this information turns out to be crucial to correctly evaluating their product. Likewise for $\infty - \infty$ and $\infty ^ 0$, which as Luboš says, are more or less the same thing (just take the $\log$ or $\exp$).
A: 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 $\frac{1}{x^2}$).
A: In the context of your limit, this can be explained by the fact that your "infinity" is also a $1/0$:
$$
\lim_{x \rightarrow 0^+} x \ln( e^{2x} -1 ) = \frac{x}{\frac1{\ln( e^{2x} -1 )}}
$$
And since as $x \rightarrow 0^+$, $\ln( e^{2x} -1 ) \rightarrow +\infty$, you get that $\frac1{\ln( e^{2x} -1 )} \rightarrow 0^+$, which means that your limit becomes $0/0$.
It's slightly more obvious why $0/0$ is indeterminate because the solution for $x=0/0$ is the solution for $0x=0$, and every number solves that.
A: 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 $0$.
A: If $f(x) \to 0$ and $g(x) \to \infty$, then the product $f(x) g(x)$ may be approaching any number at all.  For example, the product may be approaching 0:
$$
\begin{array}{c|c|c|c|c|c}
f(x) & 0.01 & 0.0001 & 0.000001 & 0.00000001 & \cdots \\
\hline
g(x) & 10 & 100 & 1000 & 10,000 & \cdots \\
\hline
f(x) g(x) & 0.1 & 0.01 & 0.001 & 0.0001 & \cdots \\
\end{array}
$$
or the product may be approaching infinity:
$$
\begin{array}{c|c|c|c|c|c}
f(x) & 0.1 & 0.01 & 0.001 & 0.0001 & \cdots \\
\hline
g(x) & 100 & 10,000 & 1,000,000 & 100,000,000 & \cdots \\
\hline
f(x) g(x) & 10 & 100 & 1000 & 10,000 & \cdots \\
\end{array}
$$
It can also approach anything in between.  For example, in the limit
$$
\lim_{x\to\infty} (x)\left(\frac{5}{x}\right)
$$
the $x$ approaches $\infty$ and the $\dfrac{5}{x}$ approaches $0$, but the product is equal to $5$.  In general, a limit of the form $0\cdot\infty$ is a competition between the two factors:


*

*If the first factor goes to $0$ more quickly, then the limit is $0$.

*If the second factor goes to $\infty$ more quickly, then the limit is $\infty$.

*If the first factor goes to $0$ at about the same rate that the second factor goes to $\infty$, then the limit may be anything in between.
You can usually solve a limit of the form $0 \cdot \infty$ using L'Hospital's rule by introducing a fraction.  Specifically, if $f(x) \to 0$ and $g(x) \to \infty$, then
$$
f(x) g(x) \;=\; \frac{f(x)}{1/g(x)}
$$
which is a fraction of the form $0/0$.  Alternatively,
$$
f(x) g(x) \;=\; \frac{g(x)}{1/f(x)}
$$
which is a fraction of the form $\infty/\infty$.  For the limit you were given the best thing is to put the $x$ in the denominator:
$$
\lim_{x\to 0^+} x\ln(e^{2x}-1) \;=\; \lim_{x\to 0^+} \frac{\ln(e^{2x}-1)}{1/x}.
$$
The fraction on the right is of the form $\infty/\infty$, so we can apply L'Hospital's rule:
$$
\lim_{x\to 0^+} \frac{\ln(e^{2x}-1)}{1/x} \;=\; \lim_{x\to 0^+} \frac{2 e^{2x} / (e^{2x}-1)}{-1/x^2}
$$
The right-hand side simplifies to
$$
\lim_{x\to 0^+} \frac{-2x^2 e^{2x}}{e^{2x} - 1}.
$$
This has the form $0/0$, so we can apply L'Hospital's rule again to get
$$
\lim_{x\to 0^+} \frac{-4x e^{2x} - 4x^2 e^{2x}}{2 e^{2x}}.
$$
This simplifies to
$$
\lim_{x\to 0^+} -2x-2x^2
$$
so the limit in this case is 0.
