1. **Problem Statement:** Find the left-hand limit, right-hand limit, and the limit at $a=2$ for the function $$f(x) = \frac{x^2 + 3x + 2}{x^2 - 4}$$ 2. **Recall the limit definitions:** - Left-hand limit: $\lim_{x \to a^-} f(x)$ means approaching $a$ from values less than $a$. - Right-hand limit: $\lim_{x \to a^+} f(x)$ means approaching $a$ from values greater than $a$. - Limit at $a$: $\lim_{x \to a} f(x)$ exists only if left and right limits are equal. 3. **Simplify the function:** Factor numerator and denominator: $$x^2 + 3x + 2 = (x+1)(x+2)$$ $$x^2 - 4 = (x-2)(x+2)$$ So, $$f(x) = \frac{(x+1)(x+2)}{(x-2)(x+2)}$$ For $x \neq -2$, this simplifies to: $$f(x) = \frac{x+1}{x-2}$$ 4. **Evaluate the limits at $a=2$: ** - Left-hand limit: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{x+1}{x-2}$$ As $x$ approaches 2 from the left, denominator $x-2$ approaches 0 negative, numerator $x+1$ approaches 3 positive. So, $$\lim_{x \to 2^-} f(x) = -\infty$$ - Right-hand limit: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{x+1}{x-2}$$ As $x$ approaches 2 from the right, denominator $x-2$ approaches 0 positive, numerator $x+1$ approaches 3 positive. So, $$\lim_{x \to 2^+} f(x) = +\infty$$ - Since left and right limits are not equal, the limit at $x=2$ does not exist: $$\lim_{x \to 2} f(x) \text{ does not exist}$$ **Final answers:** $$\lim_{x \to 2^-} f(x) = -\infty$$ $$\lim_{x \to 2^+} f(x) = +\infty$$ $$\lim_{x \to 2} f(x) \text{ does not exist}$$