Indeterminate forms - zero × inf determinate?

Question

Anything multiplied by zero is zero.but when we multiply inf. With zero we don't say zero but indeterminate.

I saw in a certain youtube video that actually in such type if problems we don't take the numbers themselves but we are approaching to that numbers, is that right ?

Secondly , 0 × ${\infty}$ is treated as finite number in many cases especially in physics {like in typical dipole problems}

Answer 1

What is called "$0$" in the indeterminate form $0\times\infty$ is, actually, a function which is infinitesimal when $x$ approaches a certain number.

And "$\infty$" is a divergent function $g$, of course.

So, by choosing $f$ anf $g$, you can obtain every possible result.

For example, if $f(x)=1/x$ and $g(x)=x$, then

$$ \lim_{x\to+\infty}f(x)g(x)=1, $$

but if $f(x)=1/x$ and $g(x)=x^2$, then

$$ \lim_{x\to+\infty}f(x)g(x)=+\infty, $$

and so on.

Answer 2

Anything multiplied by zero is zero.

This isn't always true.

There are lots of contexts where it's true, such as when doing arithmetic with real numbers or various kinds of arithmetic structures (e.g. rings) designed to have similar arithmetic.

But this isn't true in every situation. The arithmetic of the extended real numbers (the number system you get by including $+\infty$ and $-\infty$) is one example where it's not true.

The extended real numbers are not intended to have all of the familiar algebraic properties, but instead to quantify the behavior of limits (among other applications). Most operations we define on $\pm \infty$ are meant to be continuous there; for example, addition is continuous at $(+\infty) + 1$, so we can compute this limit by plugging in values:

$$ \lim_{x \to +\infty} (x + 1) = (+\infty) + 1 = +\infty $$

Multiplication, however, cannot be continuous at $0 \cdot (+\infty)$. If it were, we would have

$$ 0 \cdot (+\infty) = \lim_{x \to +\infty} \frac{1}{x} \cdot x^2 = \lim_{x \to +\infty} x = +\infty$$ $$ 0 \cdot (+\infty) = \lim_{x \to +\infty} \frac{1}{x} \cdot x = \lim_{x \to +\infty} 1 = 1$$ $$ 0 \cdot (+\infty) = \lim_{x \to +\infty} \frac{1}{x^2} \cdot x = \lim_{x \to +\infty} \frac{1}{x} = 0$$

Obviously, these can't all be true. That's why we don't define $0 \cdot (+\infty)$.

All of the "indeterminate forms" are examples where we cannot continuously define arithmetic on extended real numbers.

Answer 3

There are situations where $0\cdot \infty$ can be considered as $0$ like in measure theory. It simplifies the notation essentially.

But you can't give a universal definition for it. It would produce a lot of contradictions. For $\alpha,\beta>0$ you can consider

$$ \lim_{x\to \infty} cx^{\alpha}\cdot\frac1{x^\beta+1}=\begin{cases}0 & \alpha<\beta\\c & \alpha=\beta\\\infty & \alpha>\beta\end{cases}\text{ while }\lim_{x\to\infty}x^\alpha=\infty\text{ and }\lim_{x\to\infty}\frac1{x^\beta+1}=0 $$

Hence, if you read a book/article, you have always to check if the author uses a convention for $0\cdot\infty$ due to simplification of the notation.

Answer 4

Short answer:

Anything multiplied by zero is zero.

Anything multiplied by infinity is infinity.

So what about $0\times\infty$ ?

Without more information on what exactly are meant by this $0$ and this $\infty$, there is no answer.