Why is $0\cdot\infty$ considered indeterminate while $\frac {0}{\infty}$ is zero? Consider the following limit: $\lim\limits_{x \to a} \frac {f(x)}{g(x)}.$ 
Suppose that
$$\lim\limits_{x \to a} f(x)=0$$ $$\text{and}$$ $$\lim\limits_{x \to a} g(x)=\infty$$
Then, we have 
$$\lim\limits_{x \to a} \frac {f(x)}{g(x)} = \frac {0}{\infty} = 0.$$ 
Now, consider this following limit: $\lim\limits_{x \to a} ({f(x)}\cdot{g(x)}).$ 
Suppose that
$$\lim\limits_{x \to a} f(x)=0$$ $$\text{and}$$ $$\lim\limits_{x \to a} g(x)=\infty$$
Then, we have 
$$\lim\limits_{x \to a} ({f(x)}\cdot{g(x)}) \rightarrow {0}\cdot{\infty}, I.F.$$
Why is $\frac {0}{\infty}$ equal to zero while ${0}\cdot{\infty}$ is considered indeterminate? An argument I've seen for why $\frac {0}{\infty}$ equals zero is simply because zero divided by any number (in this case, infinity growing larger and large) is zero. That being said, we also know that zero $\it multiplied$ by any number is zero as well. Why can't the same rationale for $\frac {0}{\infty}=0$ be used for the indeterminate form ${0}\cdot{\infty}$?
 A: $0\cdot\infty$ is shorthand for very small (in magnitude) times very large (in magnitude).  These two effects compete, and we don't know what will happen because of this competition. 
On the other hand, $\frac{0}{\infty}$ is shorthand for very small divided by very large.  If you have a small thing and divide it into a large number of parts, each part is very very small.  Here the two extremes work together, both pushing the quotient toward very small (in magnitude).
A: It is absolutely not true that whenever $f(x)\to0$ and $g(x)\to\infty$ then $f(x)g(x)\to0.$
A very simple example is $f(x) = x$ and $g(x) = \dfrac 5 x.$
Here we have $\lim\limits_{x\to0+}f(x)=0$ and $\lim\limits_{x\to0+} g(x) = +\infty$ and $\lim\limits_{x\to0+} f(x)g(x) = 5.$
The reason a form is considered indeterminate is that the limit can be any of various different things depending on which functions are involved.
A: $$0.\infty =0.(1/0)=0/0$$
$$0.\infty =(1/{\infty}).\infty =\infty /\infty$$
 Thus it is indeterminate. 
A: The first example follows thusly:
$\lim_{x\to a} f(x) = 0$ means for any $\epsilon$ there the is a $\delta_1$ so that $|x-a| < \delta_1$ will yield $|f(x)-0|= |f(x)| \epsilon$.
And $\lim_{x\to a} g(x) = \infty$ means for any $M$ there is a $\delta_2$ so that $|x-z| < \delta_2$ will yield $g(x) > M$.
So for any $\epsilon > 0$ let $\delta_1$ be such that $|x-a|<\delta_1$ means $|f(x)-0|=|f(0)|< \min(1,\epsilon)$ and let $\delta_2$ be such that $|x-a| < \delta_2$ means $g(x) > \max (1,\frac 1{\epsilon})$ then
$|x-a| < \min(\delta_1, \delta_2)$ means $|f(x)|< \min(\epsilon, 1)$ and $g(x) > \max(1, \frac 1\epsilon)$ and so $0 < \frac 1{g(x)}< \min (1, \epsilon)$ and so $|\frac {f(x)}{g(x)}| < \min (1, \epsilon^2) \le \min(1, \epsilon)$.  
So $\lim_{x\to a}\frac {f(x)}{g(x)} = 0$.
.....
But nothing can be said about the $\lim_{x\to a} f(x)g(x)$.
We can find a $\delta$ so that $|f(x)| < \epsilon$ and $g(x) > \frac 1{\epsilon}$ but ... what does that say about $f(x)g(x)$.  One term has an absolute value less than $\epsilon$.  And the other has a term greater then $\frac 1{\epsilon}$.  But we don't know how much bigger and smaller and we have no idea how the product will behave.
I mean.... sure... we can try to do....we have $|f(x)g(x)| < \epsilon*g(x)$ and $\epsilon*g(x) > \epsilon*\frac 1\epsilon =1$.  We can't combine a $a < b$ and $b > c$ statements to get a $a < b > c$ statement.  $|f(x)g(x)| < \epsilon*g(x) > 1$ is both ungramatical and meaningless.
As is $|f(x)g(x)| > |f(x)|*\frac 1{\epsilon} < 1$.  Just can't get anything useful.
Showing something is greater than something that is less than something else is meaning less.  (I'm taller than something that is shorter than a housefly... so what.)  Likewise showing something is smaller than something greater than something else is equally meaningless.  (I'm shorter than something that is taller than a housefly... so what.)
A: If $\lim_{x\to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist and
$${\lim_{x \to a} f(x)}=0, ~~{\lim_{x \to a} g(x)}=\infty$$
then it is true that
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0} {\infty}=0\times\frac{1}{\infty}=0\times 0 = 0$$
However, the same logic does not hold for
$$\lim_{x \to a} f(x)g(x) = 0 \times \infty$$
consider 
$${\lim_{x \to 0^+} f(x)}={\lim_{x \to 0^+} x^2}=0, ~~{\lim_{x \to 0^+} g(x)}={\lim_{x \to 0^+} \frac{1}{x^2}}=\infty$$
where
$$\lim_{x \to 0^+} {f(x)}{g(x)} ={\lim_{x \to 0^+} }1=1$$
and then consider as a second example
$${\lim_{x \to 0^+} f(x)}={\lim_{x \to 0^+} x}=0, ~~{\lim_{x \to 0^+} g(x)}={\lim_{x \to 0^+} \frac{1}{x^2}}=\infty$$
where
$$\lim_{x \to 0^+} {f(x)}{g(x)} ={\lim_{x \to 0^+} }\frac{1}{x}=\infty$$
The value of 
$$\lim_{x \to 0^+} {f(x)}{g(x)}$$
depends on the functions we choose for $f(x)$ and $g(x)$. Therefore, we refer to this case as indeterminate.
A: $0 \cdot \infty$ is considered indeterminate because the answer depends on the rate at which the limiting values of $0$ and $\infty$ are approached by their respective functions. However, $\frac{0}{\infty}$ is not indeterminate becuase the answer does not depend upon the rate at which $0$ or $\infty$ is approached. In other words, if the function in the numerator is always getting smaller and function in the denominator is always getting larger, then the ratio between the two is always getting closer and closer to $0$ - no matter how quickly one is getting smaller or larger.
