Question

Evaluate: \[ \lim_{n \to \infty} \left(\frac{\sqrt{n^2 + 1} + n}{\sqrt[3]{n^6 + 1}}\right)^2 \]

Hint:

When limits involve square roots and cube roots with high powers of $n$, factor out the dominant term carefully to avoid power mismatches.

Answer 1

Correct Answer: 4

Step 1: Simplify the expression.
We have \[ L = \lim_{n \to \infty} \frac{(\sqrt{n^2 + 1} + n)^2}{(\sqrt[3]{n^6 + 1})^2}. \] Simplify numerator and denominator using leading terms.

Step 2: Factor out highest powers of $n$.
\[ \sqrt{n^2 + 1} = n\sqrt{1 + \frac{1}{n^2}} \approx n\left(1 + \frac{1}{2n^2}\right). \] So, \[ \sqrt{n^2 + 1} + n \approx n\left(1 + \frac{1}{2n^2}\right) + n = 2n + \frac{1}{2n}. \]

Step 3: Simplify denominator.
\[ \sqrt[3]{n^6 + 1} = n^2\sqrt[3]{1 + \frac{1}{n^6}} \approx n^2\left(1 + \frac{1}{3n^6}\right). \]

Step 4: Substitute and simplify.
\[ L = \lim_{n \to \infty} \frac{(2n + \frac{1}{2n})^2}{n^4(1 + \frac{1}{3n^6})^2} = \lim_{n \to \infty} \frac{4n^2 + 2 + \frac{1}{4n^2}}{n^4(1 + \frac{2}{3n^6})}. \] Dominant term in numerator: $4n^2$, denominator: $n^4$. Hence, \[ L = \lim_{n \to \infty} \frac{4n^2}{n^4} = 0. \] Wait, let's recheck: The exponent structure needs careful handling — note that $\sqrt[3]{n^6 + 1} = n^2$, not $n^4$.

Step 5: Correct simplification.
\[ L = \lim_{n \to \infty} \left(\frac{\sqrt{n^2 + 1} + n}{n^2}\right)^2 = \lim_{n \to \infty} \left(\frac{2n + \frac{1}{2n}}{n^2}\right)^2 = \lim_{n \to \infty} \left(\frac{2}{n} + \frac{1}{2n^3}\right)^2 = 0. \] Correction: We missed the cube root power. Actual denominator is $\sqrt[3]{n^6 + 1} = n^2(1 + \frac{1}{n^6})^{1/3} \approx n^2$. So, \[ L = \left(\frac{2n}{n^2}\right)^2 = \frac{4}{n^2} \to 0. \] Hence, \[ \boxed{0.} \]