Question

determine whether each limit exists. if a limit exists, estimate its value.
(a) lim f(x)
x→ - 3
(b) lim f(x)
x→0
(a) find the one - sided limits.
lim f(x)=
x→ - 3^-
lim f(x)=
x→ - 3^+

Explanation:

Step1: Analyze left - hand limit as $x\to - 3$

As $x$ approaches $-3$ from the left (values less than $-3$), we look at the graph. Following the curve from the left side towards $x = - 3$, we estimate the $y$-value.

Step2: Analyze right - hand limit as $x\to - 3$

As $x$ approaches $-3$ from the right (values greater than $-3$), we look at the graph. Following the curve from the right side towards $x=-3$, we estimate the $y$-value.

Step3: Determine limit as $x\to - 3$

If the left - hand limit and the right - hand limit as $x\to - 3$ are equal, then $\lim_{x\to - 3}f(x)$ exists and is equal to that common value.

Step4: Analyze limit as $x\to0$

We look at the behavior of the graph as $x$ approaches $0$. We check the left - hand limit (values less than $0$) and the right - hand limit (values greater than $0$). If they are equal, the limit as $x\to0$ exists.

From the graph:

For $\lim_{x\to - 3}$:

• $\lim_{x\to - 3^{-}}f(x)= - 4$
• $\lim_{x\to - 3^{+}}f(x)= - 4$

Since $\lim_{x\to - 3^{-}}f(x)=\lim_{x\to - 3^{+}}f(x)= - 4$, then $\lim_{x\to - 3}f(x)= - 4$

For $\lim_{x\to0}$:

As $x$ approaches $0$ from the left, the $y$-values approach $- 4$. As $x$ approaches $0$ from the right, the $y$-values approach $- 4$. So $\lim_{x\to0}f(x)= - 4$

Answer:

(a) $\lim_{x\to - 3}f(x)= - 4$
(b) $\lim_{x\to0}f(x)= - 4$
(a) $\lim_{x\to - 3^{-}}f(x)= - 4$, $\lim_{x\to - 3^{+}}f(x)= - 4$