Indeterminate form $0 \times \infty$ In below figure the longer polygon is obtained by halving the height and doubling the length. Both areas are same. I'm wondering what happens if we continue this forever. Height follows the sequence $\{\frac{1}{2^n}\}$ and length follows the sequence $\{2^n\}$. Then the area is given by 
$$\frac{1}{2^n}\times 2^n$$
which is always $1$. However the height seems to approach $0$. Doesn't this make area also $0$? I couldn't imagine how the area can remain $1$ even when the height becomes $0$. Any help?

 A: Actually, this is a good demonstration of why the form $0\times\infty$ is indeterminate. The height approaches $0$ and the base approaches $\infty$, yet the limit approaches $1$.
Suppose you originally begin with a rectangle of height $Y$ and base $1$ and proceded to create new rectangles by halving the height and doubling the base.
This would give a sequence of rectangles, each with area $Y$.
So you would have an example of $0\times\infty$ where the limit is $Y$.
A: Your confusion lies in the fact that you seem to be thinking that "$0 \times \infty = 0$" is true. However, this statements is not only false, it is nonsensical. "$\infty$" is not a real number so it doesn't follow the usual arithmetic properties; it is just a convenient symbol (whose meaning and usage often isn't explained properly). You said

I couldn't imagine how the area can remain $1$ even when the height becomes $0$.

Well, the point is that even though the height decreases to $0$, the base increases in such a way that it "cancels the decrease in height".

Consider $f(x) = x^2$ and $g(x) = \dfrac{1}{x^2}$ (for $x \neq 0$). You might be thinking that
\begin{align}
\lim_{x \to 0} (f(x) g(x)) &= \lim_{x \to 0}f(x) \cdot \lim_{x \to 0} g(x) \\
&= 0 \cdot \infty \\
&= ????????????????
\end{align}
and when you see the indeterminate form $0 \cdot \infty$, you get confused. However, this line of reasoning is complete nonsense. The problem arises in the very first step, you can't always split the limit of the product into the product of limits! You can only do this when both limits exist and are finite!
The actual limit is
\begin{align}
\lim_{x \to 0} \left(f(x) g(x) \right) &= \lim_{x \to 0} \left(x^2 \cdot \dfrac{1}{x^2} \right) \\
&= \lim_{x \to 0}(1) \\
&= 1
\end{align}
A: Perhaps it helps to think about it this way: take a rectangle and bisect it horizontally from left to right.  The larger rectangle is decomposed into two smaller rectangles, one above the other.
Take the rectangle “on top” and move it so it's “on the right” of the lower rectangle.  The resulting rectangle has the same area of the previous rectangle, because all we did was cut it into pieces and rearrange the pieces.  But it's equivalent to halving the height and doubling the width of the rectangle.
