Differentiability of a Piecewise Function

Let $v$ be the function defined by $\displaystyle v(x) = \begin{cases} \sqrt{x+6} & \text{if } x < 3 \\ 2x - 3 & \text{if } x \geq 3 \end{cases}$. Which of the following statements about $v$ is true?

I. $\displaystyle \lim_{x \to 3^-} v(x) = \lim_{x \to 3^+} v(x)$

II. $\displaystyle \lim_{x \to 3^-} v'(x) = \lim_{x \to 3^+} v'(x)$

III. $v$ is differentiable at $x = 3$.