One-sided limits and existence of a two-sided limit at a piecewise boundary

As $x \to 0^-$, we use the branch $f(x) = e^x$ since this applies for $x < 0$. Since $e^x$ is continuous, $\lim_{x \to 0^-} e^x = e^0 = 1$.

As $x \to 0^+$, we use the branch $f(x) = 2 - x^2$ since this applies for $x \geq 0$. Since this is a polynomial, $\lim_{x \to 0^+} (2 - x^2) = 2 - 0 = 2$. Since the left-hand limit ($1$) and right-hand limit ($2$) are unequal, $\lim_{x \to 0} f(x)$ does not exist.