Answer
We could guess that the value of the limit $\lim\limits_{x \to \infty}\frac{x^2}{2^x}$ is $0$
On the graph, we can see that the function approaches $0$ as $x$ becomes more positive. We can see that $\lim\limits_{x \to \infty}\frac{x^2}{2^x} = 0$
Work Step by Step
$f(x) = \frac{x^2}{2^x}$
We can evaluate $f(x)$ at various values of $x$:
$f(0) = \frac{0^2}{2^0} = 0$
$f(1) = \frac{1^2}{2^1} = 0.5$
$f(2) = \frac{2^2}{2^2} = 1$
$f(3) = \frac{3^2}{2^3} = 1.125$
$f(4) = \frac{4^2}{2^4} = 1$
$f(5) = \frac{5^2}{2^5} = 0.78125$
$f(6) = \frac{6^2}{2^6} = 0.5625$
$f(7) = \frac{7^2}{2^7} = 0.3828125$
$f(8) = \frac{8^2}{2^8} = 0.25$
$f(9) = \frac{9^2}{2^9} = 0.158203125$
$f(10) = \frac{10^2}{2^{10}} = 0.09765625$
$f(20) = \frac{20^2}{2^{20}} = 0.00038147$
$f(50) = \frac{50^2}{2^{50}} = 0.00000000000222$
$f(100) = \frac{100^2}{2^{100}} = 7.888609\times 10^{-27}$
We could guess that the value of the limit $\lim\limits_{x \to \infty}\frac{x^2}{2^x}$ is $0$
On the graph, we can see that the function approaches $0$ as $x$ becomes more positive. We can see that $\lim\limits_{x \to \infty}\frac{x^2}{2^x} = 0$
