$\pi = 0\times\infty$

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I found this equation online that lets you calculate $\pi$: “$\pi = x \times \sin(180/x)$”, where $x$ must be a “big” number (close to infinity) so that it's as accurately close to $\pi$ as possible.

Since replacing $x$ with infinity would give us the exact value of $\pi$, I came to the following conclusion:

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$\pi = \lim_{x\to\infty}x\times\sin(180/x)$

$\pi = \lim_{x\to\infty}x\times\sin(0)$

$\pi = \lim_{x\to\infty}0\times x$

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But “$0 \times\infty$” is not defined, so are my calculations incorrect?

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You can not isolate limits.

You can not do $\lim_{x\rightarrow p} f(x)g(x) = \lim_{x\rightarrow p} [f(x)*\lim_{y\rightarrow p} g(y)]$.

That simply is not true and can not be done.

You can do $\lim_{x\rightarrow p}f(x)*g(x) = \lim_{x\rightarrow p}f(x)*\lim_{x\rightarrow p}g(x)$ but only if $ \lim_{x\rightarrow p}f(x) =A$ and $\lim_{x\rightarrow p}g(x) = B$ and $A*B$ are all defined. As $0*\infty$ is not defined we can not do it here.