How are these indeterminate ?

**pratique21** · July 18th 2012, 11:03 AM

0 * infinity
1^ infinity

**emakarov** · July 18th 2012, 11:27 AM

Originally Posted by pratique21

0 * infinity
1^ infinity

I guess these are indeterminate in the sense that if $\lim_{x\to\infty} f(x)=0$ and $\lim_{x\to\infty} g(x)=\infty$ , then $\lim_{x\to\infty}f(x)g(x)$ can be any number. For example, for $a>0$ , $\lim_{x\to\infty}\frac{1}{x}\cdot ax=a$ .

Similarly, $\lim_{n\to\infty}\left(1+\frac{a}{n}\right)=1$ and $\lim_{n\to\infty}n=\infty$ , but $\lim_{n\to\infty}\left(1+\frac{a}{n}\right)^n=e^a$ .

**Soroban** · July 18th 2012, 11:51 AM

Hello, pratique21!

How are these indeterminate? . $0\cdot\infty,\;\;1^{\infty}$

I'll explain the first one . . .

Consider the function: . $f(x) \:=\:\sin x\cdot\frac{1}{x}\,\text{ as }x\to0.$
We see that: . $\lim_{x\to0}\left(\sin x \cdot\frac{1}{x}\right) \:=\:0\cdot\infty$

But we have a theorem that states: . $\lim_{x\to0}\frac{\sin x}{x} \:=\:1$

. . Hence, we could say: . $0\cdot\infty \:=\:1$

Now consider the function: . $g(x) \:=\:\sin2x\cdot\frac{1}{x}\,\text{ as }x\to0.$

We see that: . $\lim_{x\to0}\left(\sin2x\cdot\frac{1}{x}\right) \:=\:0\cdot\infty$

But: . $\lim_{x\to0}\frac{\sin2x}{x} \;=\;\lim_{x\to0} \left(2\!\cdot\!\frac{\sin2x}{2x} \right) \;=\; 2\cdot\lim_{x\to0}\frac{\sin2x}{2x} \;=\; 2\cdot1 \;=\;2$

. . Hence, we could say: . $0\cdot\infty \:=\:2$

In general, we can show that $0\cdot\infty$ can equal any number.

Therefore, $0\cdot\infty$ is indeterminate.

**pratique21** · July 18th 2012, 09:02 PM

Thank you Emakarov and soroban. I can understand it now. So that is the also why Infinity - infinity is indeterminate as well- It can take any value.

**richard1234** · July 18th 2012, 09:34 PM

@pratique, $\infty - \infty$ is also indeterminate.

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