Checking the three-part continuity test at x = 0

Since $x = 0$ satisfies $x \le 0$, we use $v(x) = e^x$ to get $v(0) = e^0 = 1$. For the limit, compute the left-hand limit using $e^x$: $\lim_{x \to 0^-} e^x = 1$. Compute the right-hand limit using $x+1$: $\lim_{x \to 0^+} (x+1) = 1$.

Since both one-sided limits equal $1$, $\lim_{x \to 0} v(x) = 1$ exists. Comparing this to $v(0) = 1$, the limit equals the function value, so all three conditions of continuity are satisfied. Therefore $v$ is continuous at $x = 0$.