1. **State the problem:** We need to explain why the function \(f(x) = \frac{1}{x+2}\) is discontinuous at \(a = -2\). 2. **Recall the definition of continuity at a point:** A function \(f(x)\) is continuous at \(x = a\) if three conditions hold: - \(f(a)\) is defined. - The limit \(\lim_{x \to a} f(x)\) exists. - The limit equals the function value: \(\lim_{x \to a} f(x) = f(a)\). 3. **Check if \(f(a)\) is defined:** Here, \(f(-2) = \frac{1}{-2 + 2} = \frac{1}{0}\), which is undefined. 4. **Check the limit \(\lim_{x \to -2} \frac{1}{x+2}\):** - As \(x \to -2^+\), \(x+2\) approaches 0 from the positive side, so \(f(x) \to +\infty\). - As \(x \to -2^-\), \(x+2\) approaches 0 from the negative side, so \(f(x) \to -\infty\). Since the left and right limits are not equal, the limit does not exist. 5. **Conclusion:** Since \(f(-2)\) is undefined and the limit \(\lim_{x \to -2} f(x)\) does not exist, the function is discontinuous at \(x = -2\). Final answer: The function \(f(x) = \frac{1}{x+2}\) is discontinuous at \(a = -2\) because \(f(-2)\) is undefined and the limit does not exist due to a vertical asymptote.