Is it true that $\pi = \lim_{x \to \infty} 10^{x+2}\sin (1.8\times 10^{-x})$? (in degrees)

Question

When I was doing experiments, I found this shocking results(!). Am I hallucinating? Why I found that the decimal digit expansions of constant $\pi$ is NOTHING MORE than the decimal digit expansions of a sine function.

I found that $$\pi = \lim_{x \to \infty} (10^{x+2}\sin (1.8\times 10^{-x})) \quad\text{(in degrees, not radians)}$$

Is this true?

Answer 1

In radians your formula reads: $$ \lim_{x\to\infty}\frac{\sin{(\pi\cdot 10^{-x-2})}}{10^{-x-2}}=\pi $$ by the well-known limit: $$ \lim_{x\to0}\frac{\sin{(ax)}}{x}=a. $$