1. **Problem:** Test the limit, continuity, and differentiability of the function $$f(x) = \begin{cases} x^2, & x < 2 \\ 5, & x = 2 \\ x + 1, & x > 2 \end{cases}$$ 2. **Step 1: Check the limit from the left at $x=2$** $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4$$ 3. **Step 2: Check the limit from the right at $x=2$** $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x + 1) = 2 + 1 = 3$$ 4. **Step 3: Check the value of the function at $x=2$** $$f(2) = 5$$ 5. **Step 4: Determine if the limit exists at $x=2$** Since $$\lim_{x \to 2^-} f(x) = 4 \neq 3 = \lim_{x \to 2^+} f(x),$$ the limit $$\lim_{x \to 2} f(x)$$ does not exist. 6. **Step 5: Check continuity at $x=2$** Continuity requires $$\lim_{x \to 2} f(x) = f(2),$$ but the limit does not exist, so $$f$$ is not continuous at $$x=2$$. 7. **Step 6: Check differentiability at $x=2$** Differentiability requires continuity, so $$f$$ is not differentiable at $$x=2$$. **Final answer:** The function $$f(x)$$ is neither continuous nor differentiable at $$x=2$$ because the left and right limits are not equal.