Question

functions f and g are graphed.
find \\(\lim\limits_{x\to 1} \dfrac{f(x)}{g(x)}\\).
choose 1 answer:
a \\(0\\)
b \\(1\\)
c \\(3\\)
d \\(4\\)
e the limit doesnt exist.

Explanation:

Step1: Find $\lim_{x\to 1} f(x)$

To find the limit of \( f(x) \) as \( x \) approaches 1, we look at the graph of \( f \). As \( x \) approaches 1 from both the left and the right, the \( y \)-value approaches 4 (the open circle at \( (1, 4) \) and the behavior around \( x = 1 \) for \( f(x) \)). So, \( \lim_{x\to 1} f(x)=4 \)? Wait, no, wait. Wait, looking at the first graph (f), at \( x = 1 \), the left side: the horizontal line is at \( y = 3 \) until \( x = 1 \), then from the right, the line goes up to the open circle at \( (1, 4) \)? Wait, no, maybe I misread. Wait, the first graph: the function \( f \) has a horizontal segment at \( y = 3 \) from \( x = -2 \) to \( x = 1 \)? Wait, no, the grid: let's check the coordinates. The first graph (f): at \( x = 0 \), \( y = 3 \). Then, as \( x \) approaches 1 from the left, the function is at \( y = 3 \)? Wait, no, the blue dot is at \( (1, 3) \), and the open circle is at \( (1, 4) \). Wait, no, maybe I messed up. Wait, the first graph: the function \( f \) has a segment from \( x = -3 \) to \( x = -2 \) going up to \( y = 4 \), then down to \( y = 3 \) at \( x = -2 \), then horizontal to \( x = 1 \) at \( y = 3 \), then from \( x = 1 \), it goes up to an open circle at \( (1, 4) \), then down. Wait, no, the blue dot is at \( (1, 3) \), and the open circle is at \( (1, 4) \). Wait, maybe the limit of \( f(x) \) as \( x \to 1 \): from the left, the function is at \( y = 3 \) (horizontal line), and from the right, it's approaching the open circle at \( y = 4 \)? No, that can't be. Wait, no, maybe the graph is different. Wait, the second graph is \( g(x) \). Wait, let's re-examine.

Wait, the problem is to find \( \lim_{x\to 1} \frac{f(x)}{g(x)} \). So we need to find \( \lim_{x\to 1} f(x) \) and \( \lim_{x\to 1} g(x) \), then use the quotient rule for limits: if \( \lim_{x\to a} g(x) eq 0 \), then \( \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)} \).

First, find \( \lim_{x\to 1} f(x) \): looking at the graph of \( f \), as \( x \) approaches 1 from the left, the function is at \( y = 3 \) (the horizontal segment), and from the right, the function approaches the open circle at \( (1, 4) \)? Wait, no, that would mean the left limit is 3 and the right limit is 4, so the limit of \( f(x) \) as \( x \to 1 \) doesn't exist? But that can't be. Wait, maybe I misread the graph. Wait, the first graph (f): the horizontal line is at \( y = 3 \) from \( x = -2 \) to \( x = 1 \), then at \( x = 1 \), there's a blue dot at \( (1, 3) \) and an open circle at \( (1, 4) \), then the function goes down. Wait, no, maybe the open circle is at \( (1, 4) \), and the blue dot is at \( (1, 3) \). So the left limit of \( f(x) \) as \( x \to 1 \) is 3 (since from the left, it's the horizontal line at \( y = 3 \)), and the right limit: as \( x \) approaches 1 from the right, the function is coming from the open circle at \( (1, 4) \)? No, that doesn't make sense. Wait, maybe the graph of \( f \) is such that as \( x \to 1 \), both left and right limits are 4? Wait, no, the blue dot is at \( (1, 3) \), which is the function's value at \( x = 1 \), but the limit is about the approach. Wait, maybe I made a mistake. Let's look at \( g(x) \).

Now, \( g(x) \): the second graph. At \( x = 1 \), the function \( g(x) \) has a horizontal segment at \( y = 1 \) from \( x = 1 \) to \( x = 2 \). So as \( x \to 1 \), from the left, \( g(x) \) is approaching \( y = 1 \) (since the segment from \( x = -1 \) to \( x = 1 \) is increasing to \( y = 1 \) at \( x = 1 \)), and from the right, it's also at \( y = 1 \). So \( \lim_{x\to 1} g(x) = 1 \).

Now, back to \( f(x) \): the first graph. Wait, maybe the limit of \( f(x) \) as \( x \to 1 \) is 4? Wait, no, the blue dot is at \( (1, 3) \), but the open circle is at \( (1, 4) \). Wait, maybe the function \( f(x) \) as \( x \) approaches 1 from the left: the horizontal line is at \( y = 3 \), and from the right, it's approaching the open circle at \( y = 4 \). But that would mean the left limit is 3 and the right limit is 4, so the limit of \( f(x) \) as \( x \to 1 \) doesn't exist. But then the limit of the quotient would not exist. But that's option E. Wait, but maybe I misread the graph.

Wait, no, maybe the first graph: the function \( f \) has a segment from \( x = -3 \) to \( x = -2 \) up to \( y = 4 \), then down to \( y = 3 \) at \( x = -2 \), then horizontal to \( x = 1 \) at \( y = 3 \), then from \( x = 1 \), it goes up to an open circle at \( (1, 4) \), then down. Wait, so as \( x \) approaches 1 from the left, \( f(x) = 3 \), and from the right, \( f(x) \) approaches 4. So \( \lim_{x\to 1^-} f(x) = 3 \), \( \lim_{x\to 1^+} f(x) = 4 \). Therefore, \( \lim_{x\to 1} f(x) \) does not exist. Then, \( \lim_{x\to 1} \frac{f(x)}{g(x)} \): since the limit of \( f(x) \) doesn't exist, does the limit of the quotient exist? Wait, but \( g(x) \) as \( x \to 1 \) has limit 1 (since from both sides, \( g(x) \) approaches 1). Wait, but if \( \lim_{x\to 1} f(x) \) doesn't exist, then the limit of the quotient might not exist. But let's check again.

Wait, maybe I messed up the graph of \( f(x) \). Let's look again: the first graph (f) has a blue dot at \( (1, 3) \) and an open circle at \( (1, 4) \). So the function value at \( x = 1 \) is 3, but the limit as \( x \to 1 \): from the left, the function is at \( y = 3 \) (horizontal line), and from the right, it's approaching the open circle at \( y = 4 \). So left limit is 3, right limit is 4. Therefore, \( \lim_{x\to 1} f(x) \) does not exist. Then, the limit of \( \frac{f(x)}{g(x)} \) as \( x \to 1 \): since the numerator's limit doesn't exist and the denominator's limit is 1 (non-zero), the limit of the quotient does not exist. Therefore, the answer is E.

Wait, but maybe I made a mistake. Let's check the options. Option E is "The limit doesn't exist." So that's the answer.

Step2: Find $\lim_{x\to 1} g(x)$

For \( g(x) \), as \( x \) approaches 1 from both the left and the right, the \( y \)-value approaches 1 (since the graph of \( g \) has a segment that reaches \( y = 1 \) at \( x = 1 \) and is horizontal around there). So \( \lim_{x\to 1} g(x) = 1 \).

Step3: Analyze the limit of the quotient

The limit of a quotient is the quotient of the limits, provided that the limit of the denominator is not zero. Here, \( \lim_{x\to 1} g(x) = 1 eq 0 \), so \( \lim_{x\to 1} \frac{f(x)}{g(x)} = \frac{\lim_{x\to 1} f(x)}{\lim_{x\to 1} g(x)} \), but only if \( \lim_{x\to 1} f(x) \) exists. However, as we saw, \( \lim_{x\to 1} f(x) \) does not exist (left limit 3, right limit 4), so the limit of the quotient does not exist.

Answer:

E. The limit doesn't exist.