## Calculus 10th Edition

Yes, the function is differentiable at $x = 2$.
Derivative from the left: $\lim\limits_{x \to 2^{-}} f(x)$ $= \lim\limits_{x \to 2^{-}} \frac{f(x) - f(c)}{x - c}$ $= \lim\limits_{x \to 2^{-}} \frac{(\frac{1}{2}x + 1) - (\frac{1}{2}(2) + 1)}{x - 2}$ $= \lim\limits_{x \to 2^{-}} \frac{(\frac{1}{2}x + 1) - 2}{x - 2}$ $= \lim\limits_{x \to 2^{-}} \frac{\frac{1}{2}x - 1}{x - 2}$ $= \lim\limits_{x \to 2^{-}} \frac{\frac{1}{2}(x - 2)}{x - 2}$ $= \lim\limits_{x \to 2^{-}} \frac{1}{2}$ $= \frac{1}{2}$ Derivative from the right: $\lim\limits_{x \to 2^{+}} f(x)$ $= \lim\limits_{x \to 2^{+}} \frac{f(x) - f(c)}{x - c}$ $= \lim\limits_{x \to 2^{+}} \frac{\sqrt {2x} - \sqrt {2(2)}}{x - 2}$ $= \lim\limits_{x \to 2^{+}} \frac{\sqrt {2x} - 2}{x - 2}$ $= \lim\limits_{x \to 2^{+}} (\frac{\sqrt {2x} - 2}{x - 2})(\frac{\sqrt {2x} + 2}{\sqrt {2x} + 2})$ $= \lim\limits_{x \to 2^{+}} \frac{2x - 4}{(x - 2)(\sqrt {2x} + 2)}$ $= \lim\limits_{x \to 2^{+}} \frac{2(x - 2)}{(x - 2)(\sqrt {2x} + 2)}$ $= \lim\limits_{x \to 2^{+}} \frac{2}{\sqrt {2x} + 2}$ $= \frac{2}{\sqrt {2(2)} + 2}$ $= \frac{2}{2 + 2}$ $= \frac{1}{2}$ Since $\lim\limits_{x \to 2^{-}} f(x) = \lim\limits_{x \to 2^{+}} f(x)$, the function $f(x)$ is differentiable at $x = 2$.