Find value(s) of \( a \) such that the following function is continuous at \( x=0 \). \[ f(x)=\left\{\begin{array}{ll}x^{2}-3 & x \leq 0 \\ x+a & x>0\end{array}\right. \]

To ensure that the function \( f(x) \) is continuous at \( x = 0 \), we need to make sure that the limit of \( f(x) \) as \( x \) approaches 0 from the left equals the limit as \( x \) approaches 0 from the right, and that both are equal to \( f(0) \). 1. First, we find \( f(0) \) using the definition for \( x \leq 0 \): \[ f(0) = 0^2 - 3 = -3. \] 2. Next, we find the left-hand limit as \( x \) approaches 0: \[ \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} (x^2 - 3) = 0^2 - 3 = -3. \] 3. Now, we find the right-hand limit as \( x \) approaches 0: \[ \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} (x + a) = 0 + a = a. \] 4. For continuity at \( x = 0 \), we need: \[ \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) = f(0). \] Thus, we set: \[ -3 = a. \] Therefore, the value of \( a \) for which \( f(x) \) is continuous at \( x = 0 \) is \[ \boxed{-3}. \]