11. for the function g graphed in the accompanying figure, find (a) $limlimits_{x \to 0^-} g(x)$ (b) $limlimits_{x \to 0^+} g(x)$ (c) $limlimits_{x \to 0} g(x)$ (d) $g(0)$ (e) $limlimits_{x \to -\infty} g(x)$ (f) $limlimits_{x \to +\infty} g(x)$. figure ex-11

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Analyze left - hand limit as $x
ightarrow0^{-}$
To find $\lim_{x
ightarrow0^{-}}G(x)$, we look at the behavior of the function $G(x)$ as $x$ approaches $0$ from the left (values of $x$ less than $0$ getting closer to $0$). From the graph, as $x$ approaches $0$ from the left, the function values approach $2$.
$\lim_{x
ightarrow0^{-}}G(x) = 2$

### Part (b)
# Explanation:
## Step1: Analyze right - hand limit as $x
ightarrow0^{+}$
To find $\lim_{x
ightarrow0^{+}}G(x)$, we look at the behavior of the function $G(x)$ as $x$ approaches $0$ from the right (values of $x$ greater than $0$ getting closer to $0$). From the graph, as $x$ approaches $0$ from the right, the function values approach $2$.
$\lim_{x
ightarrow0^{+}}G(x)=2$

### Part (c)
# Explanation:
## Step1: Recall the definition of two - sided limit
The two - sided limit $\lim_{x
ightarrow a}f(x)$ exists if and only if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)$. We know from part (a) that $\lim_{x
ightarrow0^{-}}G(x) = 2$ and from part (b) that $\lim_{x
ightarrow0^{+}}G(x)=2$. Since the left - hand limit and the right - hand limit are equal, $\lim_{x
ightarrow0}G(x) = 2$

### Part (d)
# Explanation:
## Step1: Analyze the value of $G(0)$
To find $G(0)$, we look at the graph of the function at $x = 0$. The graph has an open circle at $x = 0$, which means the function is not defined at $x = 0$ (or the value is not equal to the limit). Wait, no, looking at the graph again, the open circle might be a mis - interpretation. Wait, the graph of $G(x)$ is a periodic wave - like graph. Wait, actually, from the graph, at $x = 0$, the function has an open circle, but maybe I made a mistake. Wait, no, let's re - examine. The graph of $y = G(x)$ is a periodic function. Wait, the open circle at $x = 0$ indicates that $G(0)$ is not equal to the limit? Wait, no, maybe the graph is such that at $x = 0$, the function has a hole? Wait, no, the problem's graph: looking at the grid, the function at $x = 0$ has an open circle, but maybe the value of $G(0)$ is not defined? Wait, no, maybe I misread. Wait, the graph is a wave that repeats, and at $x = 0$, the open circle suggests that $G(0)$ is not equal to the limit, but actually, maybe the function is not defined at $x = 0$? Wait, no, the problem says "find $G(0)$". From the graph, the open circle at $x = 0$ means that the function has a discontinuity at $x = 0$, and $G(0)$ is not equal to the limit. Wait, but maybe the open circle is a mistake, or maybe the function at $x = 0$ is not defined. Wait, no, looking at the graph, the function's graph near $x = 0$: the left - hand and right - hand limits are $2$, but the point at $x = 0$ is an open circle, so $G(0)$ does not exist (or is not equal to the limit). Wait, but maybe the graph is such that $G(0)$ is undefined? Wait, no, the problem is from a calculus limit problem, usually, in such graphs, if there is an open circle at $x=a$, $f(a)$ is either undefined or has a different value. But maybe in this case, the graph is a periodic function, and at $x = 0$, the function has a hole, so $G(0)$ is not equal to the limit. Wait, but let's check again. The graph of $y = G(x)$ is a wave that repeats, with peaks at $y = 2$ and troughs at some lower value. At $x = 0$, the open circle: so $G(0)$ is not equal to the limit. Wait, but maybe the problem has a typo, or maybe I missee. Wait, no, the standard way: if there is an open circle at $x = 0$, $G(0)$ is not defined (or the value is not given by the limit). But maybe in this case, the function at $x = 0$ is undefined, but that's unlikely. Wait, no, maybe the open circle is a mistake, and the function is continuous? No, the open circle indicates a discontinuity. Wait, perhaps the answer is that $G(0)$ is not equal to the limit, but let's see the graph again. The graph is a periodic function, so as $x
ightarrow\pm\infty$, we can analyze the end - behavior. But for $G(0)$, since there is an open circle, $G(0)$ is undefined? Wait, no, maybe the open circle is at a different value. Wait, the $y$ - axis: the open circle is at $x = 0$, and the function's graph around $x = 0$ has a peak at $y = 2$ on both sides. Wait, maybe the open circle is a mistake, and $G(0)=2$? No, open circle means the function does not take that value at $x = 0$. Wait, maybe the function is defined as the wave, and at $x = 0$, it's a hole, so $G(0)$ is undefined. But that's strange. Wait, maybe I made a mistake. Let's proceed.

### Part (e)
# Explanation:
## Step1: Analyze end - behavior as $x
ightarrow-\infty$
To find $\lim_{x
ightarrow-\infty}G(x)$, we look at the behavior of the function as $x$ becomes very large in the negative direction. The graph of $G(x)$ is a periodic function (it repeats its pattern). For a periodic function, as $x
ightarrow\pm\infty$, if the function is periodic with period $T$, the limit as $x
ightarrow\pm\infty$ does not exist (oscillates) unless the function is constant. But looking at the graph, the function oscillates between a minimum and maximum value. Wait, but the graph shows a wave that repeats, so as $x
ightarrow-\infty$, the function does not approach a single value, it oscillates. Wait, no, maybe the graph is such that as $x
ightarrow-\infty$, the function's values oscillate, but maybe the problem's graph is a periodic function with a repeating pattern, so the limit as $x
ightarrow-\infty$ does not exist (oscillates). But wait, maybe I misread the graph. Wait, the left - hand side of the graph: as $x$ goes to $-\infty$, the function continues to oscillate, so $\lim_{x
ightarrow-\infty}G(x)$ does not exist (oscillates). But maybe the problem expects us to see that the function is periodic and does not have a limit as $x
ightarrow\pm\infty$.

### Part (f)
# Explanation:
## Step1: Analyze end - behavior as $x
ightarrow+\infty$
To find $\lim_{x
ightarrow+\infty}G(x)$, we look at the behavior of the function as $x$ becomes very large in the positive direction. Since the function $G(x)$ is periodic (it repeats its wave - like pattern), as $x
ightarrow+\infty$, the function does not approach a single finite value, it oscillates between its minimum and maximum values. So $\lim_{x
ightarrow+\infty}G(x)$ does not exist (oscillates).

### Final Answers:
(a) $\boldsymbol{\lim_{x
ightarrow0^{-}}G(x)=2}$

(b) $\boldsymbol{\lim_{x
ightarrow0^{+}}G(x)=2}$

(c) $\boldsymbol{\lim_{x
ightarrow0}G(x)=2}$

(d) $\boldsymbol{G(0)}$ is undefined (or there is a discontinuity, if we assume the open circle means the function does not take the value at $x = 0$; if there was a closed circle, it would be $2$, but the open circle indicates otherwise)

(e) $\boldsymbol{\lim_{x
ightarrow-\infty}G(x)}$ does not exist (oscillates)

(f) $\boldsymbol{\lim_{x
ightarrow+\infty}G(x)}$ does not exist (oscillates)

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Analyze left - hand limit as \(x

Part (b)

Step1: Analyze right - hand limit as \(x

Part (c)

Step1: Recall the definition of two - sided limit

Part (d)

Answer:

Step1: Analyze end - behavior as \(x

Final Answers: