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: Expand the numerator

$$\left(\sqrt{n^2 + 1} + n\right)^2 = n^2 + 1 + 2n\sqrt{n^2 + 1} + n^2$$

$$= 2n^2 + 1 + 2n\sqrt{n^2 + 1}$$

Step 2: Simplify $\sqrt{n^2 + 1}$

$$\sqrt{n^2 + 1} = n\sqrt{1 + \frac{1}{n^2}} \approx n\left(1 + \frac{1}{2n^2}\right) = n + \frac{1}{2n}$$ for large $n$

Step 3: Substitute back

$$2n^2 + 1 + 2n\sqrt{n^2 + 1} \approx 2n^2 + 1 + 2n\left(n + \frac{1}{2n}\right)$$

$$= 2n^2 + 1 + 2n^2 + 1 = 4n^2 + 2$$

For large $n$, dominant term is $4n^2$

Step 4: Simplify the denominator

$$\sqrt[3]{n^6 + 1} = n^2\sqrt[3]{1 + \frac{1}{n^6}} \approx n^2\left(1 + \frac{1}{3n^6}\right) = n^2 + \frac{1}{3n^4}$$

For large $n$, dominant term is $n^2$

Step 5: Calculate the limit

$$\lim_{n \to \infty} \frac{4n^2 + 2}{n^2 + \frac{1}{3n^4}} = \lim_{n \to \infty} \frac{4n^2(1 + \frac{1}{2n^2})}{n^2(1 + \frac{1}{3n^6})} = \lim_{n \to \infty} \frac{4(1 + \frac{1}{2n^2})}{1 + \frac{1}{3n^6}}$$

$$= \frac{4 \cdot 1}{1} = 4$$

Answer: 4