The mathematical expression provided in the image appears to be:

$\lim_{x \to +\infty} \left( \frac{\sqrt{x+2} - 2}{\sqrt{x+7} - 3} \right)$

Steps to Solve:

1. Substitute $x \to +\infty$: As $x$ approaches infinity, the dominant term in both the numerator and denominator is $\sqrt{x}$. We can focus on simplifying the behavior as $x$ becomes very large.

2. Simplify using the limit properties: To handle the infinity form, we multiply both the numerator and denominator by the conjugates of their respective terms to simplify the square roots:

   $\frac{\sqrt{x+2} - 2}{\sqrt{x+7} - 3} \cdot \frac{\sqrt{x+2} + 2}{\sqrt{x+2} + 2} = \frac{(x+2) - 4}{(\sqrt{x+7} - 3)(\sqrt{x+7} + 3)}$

   Simplify each part: $= \frac{x - 2}{(x+7) - 9} = \frac{x - 2}{x - 2}$

   For large $x$, this simplifies to $1$, because the lower-order terms ($-2$, $-9$) become negligible as $x \to \infty$.

Conclusion:

$\lim_{x \to +\infty} \left( \frac{\sqrt{x+2} - 2}{\sqrt{x+7} - 3} \right) = 1$

Would you like a more detailed breakdown of any step, or do you have any specific questions?

Here are five related questions to expand your understanding:

1. What are common methods for solving limits involving infinity and square roots?
2. How do conjugates help simplify expressions involving square roots?
3. What is the impact of lower-order terms when evaluating limits at infinity?
4. How would this limit change if the coefficients of $x$ were different inside the square roots?
5. How would you apply L'Hopital's Rule to limits of indeterminate forms like this one?

Tip: When evaluating limits at infinity, always check if multiplying by a conjugate or simplifying leading terms makes the limit clearer.