Answer
$$\lim_{x\to \infty}\frac{x^2}{2^x}=0$$
Work Step by Step
Given $$\lim_{x\to \infty}\frac{x^2}{2^x}$$
Then by using the following data
$f(0) = 0,\ $
$f(1)= 0.5,\ $
$f(2) = 1,$
$f(3) =1.125$
$f(4) = 1 $
$f(5 ) = 0.78125 $
$f(6 ) = 0.5625$
$f(7 ) = 0.3828125$
$f(8 ) = 0.25$
$f(9 ) = 0.158203125$
$f(10 ) = 0.09765625 $
$f(20 ) = 0.00038147 $
$f(50) = 2.22045 \times 10^{-12}$
$f(100 ) = 7.88861 \times 10^{-27}$
Then we conclude that
$$\lim_{x\to \infty}\frac{x^2}{2^x}=0$$
The graph below also supports the conclusion from the table.
