To solve the limit problem in the image, we need to find:

$\lim_{x \to \infty} \frac{x^2 f(x) - g(x) + 5}{x^2 g(x) - 2x - 1}$

given the following information: $\lim_{x \to \infty} f(x) = 2, \quad \lim_{x \to \infty} g(x) = 5.$

Step-by-step Solution:

1. Substitute the limits of $f(x)$ and $g(x)$ into the expression:

   As $x \to \infty$, since $\lim_{x \to \infty} f(x) = 2$ and $\lim_{x \to \infty} g(x) = 5$, we substitute these values into the expression. However, we need to consider how the terms involving $x$ behave.

2. Analyze the numerator:

   $x^2 f(x) - g(x) + 5 \approx x^2 \cdot 2 - 5 + 5 = 2x^2.$ As $x \to \infty$, the $x^2 f(x)$ term dominates the expression because it grows faster than the constant terms.

3. Analyze the denominator:

   $x^2 g(x) - 2x - 1 \approx x^2 \cdot 5 - 2x - 1 = 5x^2 - 2x - 1.$ Similarly, as $x \to \infty$, the $x^2 g(x)$ term (or $5x^2$) dominates the denominator since it grows much faster than the linear and constant terms.

4. Simplify the limit expression:

   Now, the expression becomes:

   $\lim_{x \to \infty} \frac{2x^2}{5x^2} = \lim_{x \to \infty} \frac{2}{5} = \frac{2}{5}.$

Conclusion:

The limit is:

$\lim_{x \to \infty} \frac{x^2 f(x) - g(x) + 5}{x^2 g(x) - 2x - 1} = \frac{2}{5}.$

Would you like further details or have any questions?

Related Questions to Explore:

1. How do we handle limits at infinity involving exponential functions?
2. What is L'Hopital's Rule, and when can it be applied to limits like this?
3. How do polynomials of different degrees behave as $x \to \infty$?
4. What techniques are used to evaluate limits involving trigonometric functions?
5. How can we determine whether a function has a horizontal asymptote?

Tip:

When evaluating limits at infinity, focus on the terms with the highest degree since they usually dominate the behavior of the expression.