Limits Properties — Question 9

Question 9

Let $\displaystyle \lim_{x \to 3} f(x) = 5$ and $\displaystyle \lim_{x \to 3} g(x) = -2$.

Evaluate the following limit using limit laws: $$\lim_{x \to 3} \left( \frac{f(x)^2 + 3g(x)}{g(x) - f(x)} \right)$$

Justify each step using properties of limits.

Question 9 - Solution

We are given: $$\lim_{x \to 3} f(x) = 5, \qquad \lim_{x \to 3} g(x) = -2$$

We need to evaluate: $$\lim_{x \to 3} \left( \frac{f(x)^2 + 3g(x)}{g(x) - f(x)} \right)$$

Step 1: Use limit laws in numerator and denominator.

Numerator: $$\lim_{x \to 3} f(x)^2 = \left( \lim_{x \to 3} f(x) \right)^2 = 5^2 = 25$$ $$\lim_{x \to 3} 3g(x) = 3 \cdot \lim_{x \to 3} g(x) = 3 \cdot (-2) = -6$$ $$\text{So, } \lim_{x \to 3} (f(x)^2 + 3g(x)) = 25 - 6 = 19$$

Denominator: $$\lim_{x \to 3} (g(x) - f(x)) = \lim_{x \to 3} g(x) - \lim_{x \to 3} f(x) = -2 - 5 = -7$$

Step 2: Final Answer

$$\lim_{x \to 3} \left( \frac{f(x)^2 + 3g(x)}{g(x) - f(x)} \right) = \frac{19}{-7} = \boxed{-\frac{19}{7}}$$