The mathematical expression in the image is:

$\lim_{{n \to \infty}} \sqrt{n + 2} \left( \sqrt{n + 3} - \sqrt{n - 4} \right)$

To solve this limit, let's simplify the expression step by step.

Step 1: Expand the terms inside the limit

We want to evaluate the limit as $n \to \infty$.

The inner part of the expression is: $\sqrt{n + 3} - \sqrt{n - 4}$ To simplify this, we can multiply and divide by the conjugate of this difference: $\frac{\left( \sqrt{n + 3} - \sqrt{n - 4} \right) \left( \sqrt{n + 3} + \sqrt{n - 4} \right)}{\left( \sqrt{n + 3} + \sqrt{n - 4} \right)}$ This gives: $\frac{(n + 3) - (n - 4)}{\sqrt{n + 3} + \sqrt{n - 4}} = \frac{7}{\sqrt{n + 3} + \sqrt{n - 4}}$

Step 2: Substitute back into the original expression

Now, the expression becomes: $\sqrt{n + 2} \cdot \frac{7}{\sqrt{n + 3} + \sqrt{n - 4}}$

Step 3: Evaluate the limit

As $n \to \infty$, the terms $\sqrt{n + 2}$, $\sqrt{n + 3}$, and $\sqrt{n - 4}$ all behave like $\sqrt{n}$. So, we can factor out $\sqrt{n}$ from the numerator and denominator: $\lim_{{n \to \infty}} \frac{7 \cdot \sqrt{n + 2}}{\sqrt{n + 3} + \sqrt{n - 4}} \approx \lim_{{n \to \infty}} \frac{7 \cdot \sqrt{n}}{2\sqrt{n}} = \frac{7}{2}$

Thus, the value of the limit is: $\boxed{\frac{7}{2}}$

Would you like more details on any part of the solution? Here are five related questions to explore further:

1. How does multiplying by the conjugate help in simplifying limits involving square roots?
2. Can this limit be solved using L'Hopital's Rule instead of algebraic manipulation?
3. How do limits behave when dealing with square root functions as $n \to \infty$?
4. What are some common techniques for simplifying expressions with square roots in calculus?
5. How does the approximation $\sqrt{n + k} \approx \sqrt{n}$ for large $n$ help in solving limits?

Tip: Multiplying by the conjugate is a powerful technique for simplifying expressions involving square roots, especially when dealing with differences of square roots.