Find the limit of a function using the squeeze theorem.

Question

Find lim x→+∞ (x^2)/(2^[x]) using the squeeze theorem.

Note: [x] denotes the flooring of x.

Answer 1

I think that 2^{x - 1} ≤ 2^[x\ ≤ 2^{x + 1} for x ≥ 2. So x²/2^{x + 1} ≤ x²/2^[x] ≤ x²/2^{x - 1}.

Now lim x→∞ x²/2^{x + 1} = lim x→∞ x²/2^{x - 1} = 0

To prove this we can use Lopital rule 2 times.

So by Squise Theorem lim x → ∞ x²/2^[x] = 0

Answer 2

For x >= 14, we can show that 1^x < x^2 < 1.5^[x], as 1^[14] = 1 < x^2 for x > 1 and 1.5^14 = 291.93 > 15^2 (these are the values for x^2 and 1.5^[x] as x -> 15^-, that is, the left limit of 15, and the x^2 ratio to the next integer value of x is x+1^2/x^2 < 1.5 for x >= 14, and this is the ratio for 1.5^[x].

Then, the lim x -> ∞ 1^[x]/2^[x] = lim x -> ∞ .5^[x] = 0, and lim x -> ∞ 1.5^[x]/2^[x] = lim x -> ∞ .75^[x] = 0

Thus, by the squeeze theorem, as, for x >= 14, 1^[x]/2^[x] < x^2/2^[x] < 1.5^[x]/2^[x] and

lim x->∞ 1^[x]/2^[x] = lim x->∞ 1.5^[x]/2^[x] = 0, lim x->∞ x^2/2^[x] = 0.