Anonymous

For x ≥ 10 , x^3 < 2^x . (Prove using induction.) So (2^x + x^3) / (x^2 + 3^x) < 2 * 2^x / (x^2 + 3^x) < 2 * 2^x / 3^x , since x^2 > 0 = 2 * (2/3)^x , which decreases to zero as x tends to infinity. So the given expression has the same limit, 0 .

Anonymous

As x tends to infinity, you can ignore x³ and x² terms. The ratio becomes 2^x /3^x. Since 2^x is smaller than 3^x. Therefore, the limit is 0.

Anonymous

in 2^x+x^3, 2^x dominates for large x. similarly in x^2+3^x, 3^x dominates for large x as we approach infinity this is (2/3)^x with a limit of 0.

Anonymous

Answer E See my Graph https://www.desmos.com/calculator/onxock...