The limit problem in the image is as follows:

$\lim \frac{\sqrt{n^5 + 3} - \sqrt{n - 3}}{\sqrt[5]{n + 3} + \sqrt{n + 3}}$

Let's solve this limit as $n \to \infty$.

Step 1: Analyze the numerator and denominator for large $n$

• For the numerator: $\sqrt{n^5 + 3} - \sqrt{n - 3}$

  • When $n$ is large, $n^5 + 3$ behaves like $n^5$, and $n - 3$ behaves like $n$.
  • So, we approximate: $\sqrt{n^5 + 3} \approx \sqrt{n^5} = n^{5/2}, \quad \sqrt{n - 3} \approx \sqrt{n}$

  The numerator approximately becomes: $n^{5/2} - \sqrt{n}$

• For the denominator: $\sqrt[5]{n + 3} + \sqrt{n + 3}$

  • When $n$ is large, $n + 3$ behaves like $n$, so we approximate: $\sqrt[5]{n + 3} \approx n^{1/5}, \quad \sqrt{n + 3} \approx \sqrt{n}$

  The denominator approximately becomes: $n^{1/5} + \sqrt{n}$

Step 2: Simplify the limit expression

Now the expression looks like: $\lim_{n \to \infty} \frac{n^{5/2} - \sqrt{n}}{n^{1/5} + \sqrt{n}}$

• In the numerator, the term $n^{5/2}$ dominates over $\sqrt{n}$ as $n \to \infty$.
• In the denominator, the term $\sqrt{n}$ dominates over $n^{1/5}$.

Thus, the limit becomes: $\lim_{n \to \infty} \frac{n^{5/2}}{\sqrt{n}} = \lim_{n \to \infty} n^{5/2 - 1/2} = \lim_{n \to \infty} n^2$

As $n \to \infty$, $n^2 \to \infty$, so the limit diverges to infinity.

Conclusion:

The limit does not converge to a finite value; instead, it goes to infinity.

Would you like more details, or have any questions about the steps?

Here are 5 related questions you could explore:

1. How does the behavior of different power functions impact limits as $n \to \infty$?
2. Can L'Hopital's Rule be applied to limits of similar forms?
3. What happens when the numerator and denominator grow at the same rate?
4. How do we handle limits involving roots of polynomials with varying degrees?
5. Can rationalization help simplify limits involving square roots?

Tip: When working with limits at infinity, always identify which terms dominate as $n$ grows large.