Question

function g is graphed.

find \\(\lim\limits_{x\to 0} \bigl(2g(x)\bigr)\\).

choose 1 answer:

a 0

b 2

c 3

d 6

e the limit doesnt exist.

Explanation:

Step1: Recall the limit property

The limit of a constant multiple of a function is the constant multiple of the limit, i.e., $\lim_{x \to a} (c \cdot f(x)) = c \cdot \lim_{x \to a} f(x)$ where $c$ is a constant. So, $\lim_{x \to 0} (2g(x)) = 2 \cdot \lim_{x \to 0} g(x)$.

Step2: Find $\lim_{x \to 0} g(x)$

To find the limit of $g(x)$ as $x$ approaches 0, we check the left - hand limit and the right - hand limit.
- **Left - hand limit ($x \to 0^-$)**: As $x$ approaches 0 from the left (values less than 0), we look at the part of the graph for $x<0$. The line approaching $x = 0$ from the left has a $y$ - value approaching 1? Wait, no, let's re - examine the graph. Wait, when $x$ approaches 0 from the left, the graph of $g(x)$ (the line) has a $y$ - value approaching 1? Wait, no, the open circle at $x = 0$ for the left - hand part is at $y = 1$? Wait, no, looking at the graph: the left - hand part (for $x<0$) is a line that goes from $(-2,-1)$? Wait, no, the graph: when $x$ is less than 0, the function has a line that passes through $(-1,0)$ and has an open circle at $(0,1)$. Wait, no, let's look again. The left - hand side (x < 0) of the graph: the line goes from $x=-2$ (where $y = - 1$) up to $x = 0$ with an open circle at $(0,1)$. Wait, no, the right - hand side (x>0) of the graph: at $x = 0$, the closed circle is at $(0,3)$? Wait, no, the graph: the blue graph has a closed circle at $(0,3)$ and an open circle at $(0,1)$. So, when $x$ approaches 0 from the left ($x\to0^-$), we follow the graph for $x < 0$. The line for $x < 0$ has a slope. Let's calculate the slope of the left - hand line. The points: when $x=-1$, $y = 0$; when $x = 0$, the open circle is at $y = 1$. So the slope $m=\frac{1 - 0}{0-(-1)}=1$. So the equation of the left - hand line is $y-0 = 1\cdot(x + 1)$, so $y=x + 1$. When $x\to0^-$, $y=x + 1\to0 + 1=1$.
- **Right - hand limit ($x \to 0^+$)**: As $x$ approaches 0 from the right (values greater than 0), we look at the part of the graph for $x>0$. The graph for $x>0$ starts at the closed circle $(0,3)$ and goes up to $(1,4)$? Wait, no, the closed circle at $x = 0$ is at $y = 3$. So when $x$ approaches 0 from the right, the $y$ - value approaches 3? Wait, this can't be. Wait, there is a mistake here. Wait, the graph: the left - hand part (x < 0) has an open circle at $(0,1)$ and the right - hand part (x>0) has a closed circle at $(0,3)$? But for the limit to exist, the left - hand limit and the right - hand limit must be equal. Wait, no, I must have misread the graph. Wait, the problem's graph: the function $g(x)$: for $x\leq0$? No, the closed circle is at $(0,3)$ and the open circle is at $(0,1)$. Wait, no, let's look at the graph again. The blue graph: at $x = 0$, there is a closed circle at $(0,3)$ (so $g(0)=3$) and an open circle at $(0,1)$. The left - hand side (x < 0) is a line that goes from $x=-2$ (where $y=-1$) up to $x = 0$ with an open circle at $(0,1)$. The right - hand side (x>0) is a line that starts at $(0,3)$ (closed circle) and goes up to $(1,4)$ and then down. Wait, this is a contradiction. Wait, no, the key for the limit as $x\to0$: the limit of $g(x)$ as $x\to0$ is determined by the left - hand limit and the right - hand limit. Wait, no, I think I made a mistake. Let's check the limit of $g(x)$ as $x\to0$. Wait, the left - hand limit (x→0⁻): when we approach 0 from the left, the graph of $g(x)$ (the line for x < 0) has a y - value approaching 1 (the open circle at (0,1)). The right - hand limit (x→0⁺): when we approach 0 from the right, the graph of $g(x)$ (the line for x>0) has a y - value approaching 3 (the closed circle at (0,3))? Wait, no, that can't be. Wait, no, the closed circle is at (0,3) and the open circle is at (0,1). So the left - hand limit (x→0⁻) is 1 and the right - hand limit (x→0⁺) is 3? But that would mean the limit doesn't exist. But wait, no, maybe I misread the graph. Wait, the graph: the blue line for x>0 starts at (0,3) (closed circle) and goes up to (1,4), then down to (3,2). The blue line for x < 0: starts at x=-3 (y=-1) up to x = 0 with an open circle at (0,1). Wait, but that would mean $\lim_{x\to0^-}g(x)=1$ and $\lim_{x\to0^+}g(x)=3$. But that would imply that $\lim_{x\to0}g(x)$ does not exist. But wait, the problem is to find $\lim_{x\to0}(2g(x))$. Wait, no, maybe I made a mistake in the left - hand and right - hand limits. Wait, let's look at the graph again. Wait, the open circle at (0,1) and the closed circle at (0,3). Wait, no, perhaps the left - hand limit and the right - hand limit are both equal to 3? No, the left - hand line (x < 0) has an open circle at (0,1), so when x approaches 0 from the left, we follow the line for x < 0, which approaches 1. The right - hand line (x>0) starts at (0,3) (closed circle), so when x approaches 0 from the right, we follow the line for x>0, which approaches 3. Since $\lim_{x\to0^-}g(x)
eq\lim_{x\to0^+}g(x)$, $\lim_{x\to0}g(x)$ does not exist. Then $\lim_{x\to0}(2g(x))$ also does not exist? But that contradicts. Wait, no, maybe I misread the graph. Wait, the closed circle is at (0,3) and the open circle is at (0,1). Wait, maybe the left - hand limit is 3? No, the left - hand line is below the closed circle. Wait, let's check the coordinates again. The grid: each square is 1 unit. The left - hand part: the line goes from x=-2 (y=-1) to x = 0, with an open circle at (0,1). The right - hand part: the line starts at (0,3) (closed circle) and goes to (1,4), then to (3,2). So, when x approaches 0 from the left (x→0⁻), the y - value approaches 1 (from the left - hand line). When x approaches 0 from the right (x→0⁺), the y - value approaches 3 (from the right - hand line). Since the left - hand limit ($\lim_{x\to0^-}g(x)=1$) and the right - hand limit ($\lim_{x\to0^+}g(x)=3$) are not equal, $\lim_{x\to0}g(x)$ does not exist. Then, by the limit property $\lim_{x\to a}(c\cdot f(x))=c\cdot\lim_{x\to a}f(x)$ (when $\lim_{x\to a}f(x)$ exists), since $\lim_{x\to0}g(x)$ does not exist, $\lim_{x\to0}(2g(x))$ also does not exist? But wait, maybe I made a mistake. Wait, no, the closed circle is at (0,3) and the open circle is at (0,1). Wait, perhaps the left - hand limit is 3? No, the left - hand line is a different line. Wait, maybe the graph is such that when x approaches 0 from both sides, the limit is 3? No, the open circle is at (0,1) for the left - hand line. Wait, I think I messed up the left - hand and right - hand sides. Wait, the function: for x < 0, the graph is the line with the open circle at (0,1), and for x>0, the graph is the line with the closed circle at (0,3). So, $\lim_{x\to0^-}g(x)=1$ and $\lim_{x\to0^+}g(x)=3$. Since $1
eq3$, $\lim_{x\to0}g(x)$ does not exist. Therefore, $\lim_{x\to0}(2g(x))$ does not exist.

Wait, but that's option E. But let's re - check. Wait, maybe the left - hand limit is 3? No, the closed circle is at (0,3), but the left - hand line has an open circle at (0,1). Wait, maybe the graph is drawn incorrectly in my mind. Wait, the problem says "Function g is graphed". The graph has a closed circle at (0,3) and an open circle at (0,1). The left - hand side (x < 0) is a line that goes from x=-2 (y=-1) to x = 0 with an open circle at (0,1). The right - hand side (x>0) is a line that starts at (0,3) (closed circle) and goes to (1,4), then to (3,2). So, when x approaches 0 from the left, we follow the left - hand line, so y approaches 1. When x approaches 0 from the right, we follow the right - hand line, so y approaches 3. Since the left - hand limit and right - hand limit are not equal, the limit of g(x) as x approaches 0 does not exist. Then, by the scalar multiple property of limits, if $\lim_{x\to a}g(x)$ does not exist, then $\lim_{x\to a}(cg(x))$ (where c is a non - zero constant) also does not exist. So $\lim_{x\to0}(2g(x))$ does not exist.

Answer:

E. The limit doesn't exist.