1. **Problem Statement:** Determine the values of $a$ for which $\lim_{x \to a} f(x)$ exists for the piecewise function: $$f(x) = \begin{cases} e^x & \text{if } x \leq 0 \\ x - 1 & \text{if } 0 < x < 1 \\ \ln x & \text{if } x \geq 1 \end{cases}$$ 2. **Recall the limit existence rule:** The limit $\lim_{x \to a} f(x)$ exists if and only if the left-hand limit $\lim_{x \to a^-} f(x)$ and the right-hand limit $\lim_{x \to a^+} f(x)$ both exist and are equal. 3. **Check continuity and limits at the boundary points $x=0$ and $x=1$: ** - At $x=0$: - Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^x = e^0 = 1$ - Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x - 1) = 0 - 1 = -1$ - Since $1 \neq -1$, the limit at $x=0$ does not exist. - At $x=1$: - Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x - 1) = 1 - 1 = 0$ - Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \ln x = \ln 1 = 0$ - Since both limits equal 0, the limit at $x=1$ exists and equals 0. 4. **Check limits at other points:** - For $x < 0$, $f(x) = e^x$ is continuous everywhere, so limits exist for all $a < 0$. - For $0 < x < 1$, $f(x) = x - 1$ is continuous, so limits exist for all $a$ in $(0,1)$. - For $x > 1$, $f(x) = \ln x$ is continuous for $x > 0$, so limits exist for all $a > 1$. 5. **Summary:** - The limit exists for all $a \neq 0$. - At $a=0$, the limit does not exist due to a jump discontinuity. **Final answer:** $$\lim_{x \to a} f(x) \text{ exists for all } a \in (-\infty, 0) \cup (0, \infty)$$ and does not exist at $a=0$.