explain why ${\left(\frac{{1}}{{2}}\right)}^{\infty}=0$ Mathematica shows 
${\left(\frac{{1}}{{2}}\right)}^{\infty}=0$, anyone can explain why ?
I know we can get $\lim\limits_{{{x}\to\infty}}{\left(\frac{{1}}{{2}}\right)}^{{x}}={0}$ by taking limit , does ${\left(\frac{{1}}{{2}}\right)}^{\infty}$ is just an abbreviated expression of $\lim\limits_{{{x}\to\infty}}{\left(\frac{{1}}{{2}}\right)}^{{x}}$ 
P.S. $\infty = +\infty$ here.
 A: $(1/2)^2=1/4$ and $(1/2)^{10}=1/1024$ and then $(1/2)^{20}=1/1048576$. As the power gets larger, the denominator approaches infinity. This makes the value of the fraction tend to zero. 
This might help. 
A: I think this is just a definition from Wolfram Alpha. As you see here!
If you type in $\left(\frac{1}{2}\right)^\infty == \lim\limits_{x \to \infty} \left(\frac{1}{2}\right)^x $, then Wolfram Alpha give you the result: True.
A: What you mean is probably $$\lim_{x\rightarrow \infty} {\left(\frac{1}{2}\right)}^{x}$$
A limit can be defined with epsilon and delta. I.e. can we find any number $\delta$ for every $\epsilon > 0$ which the expression is close enough to.
A variant of this: We can take the binary number $2^{-i}$ for an integer $i$ which is bounded below by 0, because positive numbers are closed under multiplication. We see that we can make the expression smaller than any such binary number, by just making $x$ large enough, say $x = i + 1$. But the binary number system is good enough to express any positive number if enough digits are allowed. Therefore the limit must be 0.
