Math riddle for true experts

[Sroly]

5-3
)5^0.5-3^0.5)(5^0.5+3^0.5)=5-3
(5^0.25-3^0.25)(5^0.25+3^0.25) (5^0.5+3^0.5)=5-3
(5^1/∞-3^1/∞)(5^1/∞+3^1/∞)….(5^0.25+3^0.25)(5^0.25+3^0.25) (5^0.5+3^0.5)=5-3
(5^1/∞-3^1/∞)=5⁰-3⁰=1-1=0
5-3=0
????

 

[JeffM]

5-3
)5^0.5-3^0.5)(5^0.5+3^0.5)=5-3 True
(5^0.25-3^0.25)(5^0.25+3^0.25) (5^0.5+3^0.5)=5-3 True
(5^1/∞-3^1/∞)(5^1/∞+3^1/∞)….(5^0.25+3^0.25)(5^0.25+3^0.25) (5^0.5+3^0.5)=5-3
(5^1/∞-3^1/∞)=5⁰-3⁰=1-1=0 Now you have gone off the rails by treating infinity as a finite number. See below.
5-3=0
????

$\displaystyle \displaystyle n \in \mathbb Z\ and\ n > 0 \implies \left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) * \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right) = 5 - 3 = 2.$

$\displaystyle \displaystyle So\ \lim_{n \rightarrow \infty}\left\{\left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) * \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right)\right\} = \lim_{n \rightarrow \infty} (2) = 2.$

But you say, and truly, $\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) = 0\ and\ 0\ times\ any\ finite\ number = 0 \ne 2.$

However, $\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left\{\left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) * \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right)\right\} = \lim_{n \rightarrow \infty}\left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) * \lim_{n \rightarrow \infty} \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right).$

And $\displaystyle \displaystyle\lim_{n \rightarrow \infty} \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right) = \infty.$

$\displaystyle 0 * \infty = 2$ is true. You are not multiplying 0 times a finite quantity.

How do we know that $\displaystyle \displaystyle\lim_{n \rightarrow \infty} \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right) = \infty.$

$\displaystyle 5^{(1/2)^i} > 1 < 3^{(1/2)^i} \implies 5^{(1/2)^i} + 3^{(1/2)^i} > 2 \implies$

$\displaystyle \displaystyle \prod_{i = 1}^n\left( 5^{(1/n)^i} + 3^{(1/n)^i}\right) > 2^n \implies$

$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \prod_{i = 1}^n\left( 5^{(1/n)^i} + 3^{(1/n)^i}\right) \ge \lim_{n \rightarrow \infty}2^n = \infty.$