The image you provided contains the following limits to evaluate:

Given: $\lim_{x \to +\infty} f(x) = -5, \quad \lim_{x \to +\infty} g(x) = -1, \quad \lim_{x \to +\infty} h(x) = -5,$ we need to find the limits for the following expressions:

(a) $\lim_{x \to +\infty} \left[ f(x) + 8g(x) \right]$

(b) $\lim_{x \to +\infty} \left[ f(x)g(x) \right]$

(c) $\lim_{x \to +\infty} \frac{7}{g(x)}$

(d) $\lim_{x \to +\infty} \frac{6h(x) + 5}{x^2}$

Solution:

(a) Using the sum rule for limits: $\lim_{x \to +\infty} \left[ f(x) + 8g(x) \right] = \lim_{x \to +\infty} f(x) + 8 \cdot \lim_{x \to +\infty} g(x)$ Substitute the given limits: $= -5 + 8(-1) = -5 - 8 = -13$

(b) Using the product rule for limits: $\lim_{x \to +\infty} \left[ f(x)g(x) \right] = \lim_{x \to +\infty} f(x) \cdot \lim_{x \to +\infty} g(x)$ Substitute the given limits: $= (-5) \cdot (-1) = 5$

(c) Using the limit rule for rational functions: $\lim_{x \to +\infty} \frac{7}{g(x)} = \frac{7}{\lim_{x \to +\infty} g(x)}$ Since $\lim_{x \to +\infty} g(x) = -1$, we get: $\frac{7}{-1} = -7$

(d) Evaluate the limit: $\lim_{x \to +\infty} \frac{6h(x) + 5}{x^2} = \lim_{x \to +\infty} \frac{6(-5) + 5}{x^2} = \lim_{x \to +\infty} \frac{-30 + 5}{x^2} = \lim_{x \to +\infty} \frac{-25}{x^2}$ As $x^2$ tends to infinity, $\frac{-25}{x^2} \to 0$. Therefore: $\lim_{x \to +\infty} \frac{-25}{x^2} = 0$

Final Answers:

(a) $-13$
(b) $5$
(c) $-7$
(d) $0$

Would you like more details or have any questions?

Here are 5 related questions:

1. How does the sum rule for limits work?
2. Can you apply the product rule for limits to any two functions?
3. Why does the limit of $\frac{-25}{x^2}$ approach zero as $x$ approaches infinity?
4. What happens if $g(x)$ tends to zero in part (c)?
5. What other limit properties can be applied to more complex functions?

Tip: Always check for indeterminate forms (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$) when calculating limits.