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!