How to prove that limit (x^2)/(2^x) = 0 as x approaches infinity

I actually know HOW to prove that, but I'm stuck at solving the inequality (x^2)/(2^x) < ε for x. Any hint would help ;)

Proving it this way sounds painful.

Try this:
Let a(n) = n^2/2^n.

Then for n > 6, we have
a(n + 1) / a(n)
= ((n + 1)^2 / 2^(n + 1)) / (n^2 / 2^n)
= (n^2 + 2n + 1) / (2n^2)
= 1/2 + 1/n + 1/2n^2
< 1/2 + 1/6 + 1/72
< 3/4.

So for n > 6,
a(n) < a(6) * (3/4)^(n-6) [via repeated use of the inequality just established].
......= (36/64) * (3/4)^(n-6).

Since lim (3/4)^n = 0 [because |3/4| < 1], the result now follows from the Squeeze Theorem.

I hope this helps!

Using L'Hospital's theorem,
lim (x → ∞) x^2/2^x
= lim (x → ∞) 2x / (2^x * ln2)
= lim (x → ∞) 2 / [2^x * (ln2)^2]
= 0.