$\lim_{x\to\infty}2^{-x}$ ( epsilon-delta)

Question

I want to prove that $\displaystyle\lim_{x\to\infty}{2^{-x}}=0$.
$\ln(2^{-x})=-x\ln(2)$ so, I define $M=-\frac{\ln(\epsilon)}{\ln(2)}$
Then,
$x>M=-\frac{\ln(\epsilon)}{\ln(2)}$
$\therefore -x\ln(2)\lt\ln(\epsilon)$
$\therefore|2^{-x}|<\epsilon$

Is this proof right?