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The mathematical expression in the image involves a limit. Here is what I can interpret from it:

\[
\lim_{n \to \infty} \frac{n\sqrt{n+5} - 3\sqrt{n}}{\sqrt{n^3 - 2} + \sqrt{n^3}}.
\]

To solve this limit as \( n \to \infty \), let's break it down step by step:

### Step 1: Simplify the expression
1. **Numerator**: \( n\sqrt{n+5} - 3\sqrt{n} \)

   We can rewrite \( n\sqrt{n+5} \) as \( n\sqrt{n(1 + \frac{5}{n})} = n\sqrt{n}\sqrt{1 + \frac{5}{n}} \).

   For large \( n \), \( \sqrt{1 + \frac{5}{n}} \approx 1 + \frac{5}{2n} \) (using a first-order binomial expansion).

   Therefore:
   \[
   n\sqrt{n+5} \approx n\sqrt{n} \left(1 + \frac{5}{2n}\right) = n\sqrt{n} + \frac{5\sqrt{n}}{2}.
   \]

   So, the numerator becomes:
   \[
   n\sqrt{n+5} - 3\sqrt{n} \approx n\sqrt{n} + \frac{5\sqrt{n}}{2} - 3\sqrt{n} = n\sqrt{n} - \frac{\sqrt{n}}{2}.
   \]

### Step 2: Simplify the denominator
2. **Denominator**: \( \sqrt{n^3 - 2} + \sqrt{n^3} \)

   For large \( n \), \( \sqrt{n^3 - 2} \approx \sqrt{n^3} = n^{3/2} \), because the term \( -2 \) becomes negligible as \( n \to \infty \).

   Therefore:
   \[
   \sqrt{n^3 - 2} + \sqrt{n^3} \approx 2n^{3/2}.
   \]

### Step 3: Compute the limit
Now, substitute the simplified expressions back into the limit:

\[
\lim_{n \to \infty} \frac{n\sqrt{n} - \frac{\sqrt{n}}{2}}{2n^{3/2}}.
\]

Factor \( \sqrt{n} \) from the numerator:

\[
\lim_{n \to \infty} \frac{\sqrt{n} \left(n - \frac{1}{2}\right)}{2n^{3/2}} = \lim_{n \to \infty} \frac{n - \frac{1}{2}}{2n}.
\]

As \( n \to \infty \), \( \frac{n - \frac{1}{2}}{2n} \to \frac{1}{2} \).

### Final Answer:
\[
\boxed{\frac{1}{2}}.
\]

Would you like further details on any specific steps or have questions about this solution?

Here are 5 related questions for practice:
1. How would the result change if the denominator had different terms, like \( \sqrt{n^3 + 10} \)?
2. Can you explain why the binomial expansion is useful for approximating square roots at large \( n \)?
3. What happens if the powers of \( n \) in the numerator and denominator were different?
4. How do we justify neglecting small terms like \( -2 \) in large \( n \) approximations?
5. Can L'Hopital's Rule be applied here, and if so, how?

**Tip:** When dealing with limits at infinity, identifying dominant terms (those with the highest power of \( n \)) helps simplify the calculation!

What is the limit of (n√(n+5) - 3√n) / (√(n³ - 2) + √n³) as n tends to infinity?

Discover how to solve the limit of (n√(n+5) - 3√n) / (√(n³ - 2) + √n³) as n tends to infinity. This calculus problem involves asymptotic analysis, binomial expansion, and understanding dominant terms for large n. Suitable for students studying undergraduate calculus.

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