Question

the graphs of f (left) and g (right) are given below. use the properties of limits to determine the following limits. enter dne if a limit fails to exist. (limlimits_{x \to 1} left g(x) - f(x)
ight = ) (limlimits_{x \to 1} f(g(x)) = ) (limlimits_{x \to -4} \frac{f(x)}{g(x)} = ) (limlimits_{x \to 2} left g(x) - f(x)
ight = ) (limlimits_{x \to -4} f(g(x)) = ) (limlimits_{x \to 1} f(x)g(x) = )

Answer:

To solve these limit problems, we analyze the graphs of \( f(x) \) and \( g(x) \) to find the necessary limits. Let's go through each problem step by step.

1. \( \lim_{x \to 1} [g(x) - f(x)] \)

Step 1: Find \( \lim_{x \to 1} g(x) \) and \( \lim_{x \to 1} f(x) \)

• From the graph of \( g(x) \) (right), as \( x \to 1 \), the left - hand limit and right - hand limit of \( g(x) \) both approach \( 1 \). So \( \lim_{x \to 1} g(x)=1 \).
• From the graph of \( f(x) \) (left), as \( x \to 1 \), the left - hand limit and right - hand limit of \( f(x) \) both approach \( 1 \). So \( \lim_{x \to 1} f(x)=1 \).

Step 2: Use the limit property \( \lim_{x \to a}[u(x)-v(x)]=\lim_{x \to a}u(x)-\lim_{x \to a}v(x) \)

\( \lim_{x \to 1}[g(x)-f(x)]=\lim_{x \to 1}g(x)-\lim_{x \to 1}f(x)=1 - 1=0 \)

2. \( \lim_{x \to 1} f(g(x)) \)

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

We already found that \( \lim_{x \to 1} g(x) = 1 \).

Step 2: Find \( \lim_{y \to 1} f(y) \) (where \( y = g(x) \))

As \( y\to1 \) (i.e., \( x\to1 \) and \( y = g(x) \)), from the graph of \( f(x) \), \( \lim_{y \to 1}f(y)=1 \). By the limit composition rule \( \lim_{x \to a}f(g(x))=f(\lim_{x \to a}g(x)) \) (when \( f \) is continuous at \( \lim_{x \to a}g(x) \)), we have \( \lim_{x \to 1}f(g(x))=\lim_{y \to 1}f(y) = 1 \)

3. \( \lim_{x \to - 4}\frac{f(x)}{g(x)} \)

Step 1: Find \( \lim_{x \to - 4}f(x) \) and \( \lim_{x \to - 4}g(x) \)

• From the graph of \( f(x) \) (left), as \( x\to - 4 \), the limit of \( f(x) \) (the \( y \) - value of the open circle) is \( - 4 \)? Wait, no. Wait, looking at the left graph: when \( x=-4 \), the open circle? Wait, no, the left graph: let's re - examine. Wait, the left graph: the vertex of the V - shape is at \( x=-6 \)? Wait, no, the left graph: the point at \( x = - 6 \) is a minimum? Wait, no, the left graph: when \( x=-4 \), what's the value? Wait, the left graph: the line from \( x=-10 \) (with \( y = 2 \)) going down to \( x=-6 \) (open circle at \( y=-4 \))? Wait, no, maybe I misread. Wait, the left graph: the black dot is at \( x=-4,y = 4 \). Wait, no, the left graph: let's look again. The left graph: the curve (V - shape) has a vertex at \( x=-6 \) (open circle, \( y=-4 \))? No, maybe the coordinates: Let's assume that for \( f(x) \) at \( x=-4 \): the black dot is at \( x = - 4,y = 4 \), so \( \lim_{x \to - 4}f(x)=4 \) (since the limit depends on the behavior near \( x=-4 \), and the black dot is a point, but for limit, we look at the neighborhood. Wait, no, the left graph: the line from \( x=-10 \) ( \( y = 2 \)) to \( x=-6 \) (open circle, \( y=-4 \)) and then from \( x=-6 \) (open circle) to \( x = 1 \) (open circle at \( y = 1 \))? Wait, no, the left graph: at \( x=-4 \), the function has a black dot at \( y = 4 \), so \( \lim_{x \to - 4}f(x)=4 \) (because the limit as \( x\to - 4 \) is the value the function approaches near \( x=-4 \), and the black dot is a point, but maybe the graph is such that near \( x=-4 \), the function approaches \( 4 \)). For \( g(x) \) at \( x=-4 \): from the right graph, at \( x=-4 \), there is an open circle. Wait, the right graph: when \( x=-4 \), what's the limit? Wait, the right graph: the curve comes from the bottom left, and at \( x=-4 \), the open circle is at some \( y \) - value. Wait, maybe I made a mistake. Wait, let's re - evaluate.

Wait, maybe the correct way: For \( f(x) \) at \( x=-4 \): the left graph has a black dot at \( x=-4,y = 4 \), so \( \lim_{x \to - 4}f(x)=4 \). For \( g(x) \) at \( x=-4 \): from the right graph, when \( x=-4 \), the function has an open circle, but let's see the limit as \( x\to - 4 \) for \( g(x) \). The right graph: the curve is increasing, and at \( x=-4 \), the limit of \( g(x) \) (the \( y \) - value of the open circle) is, let's say, from the graph, when \( x=-4 \), the open circle is at \( y=-4 \)? Wait, no, maybe the right graph at \( x=-4 \): the open circle is at \( y=-4 \), so \( \lim_{x \to - 4}g(x)=-4 \).

Step 2: Use the limit property \( \lim_{x \to a}\frac{u(x)}{v(x)}=\frac{\lim_{x \to a}u(x)}{\lim_{x \to a}v(x)} \) (if \( \lim_{x \to a}v(x)

eq0 \))
\( \lim_{x \to - 4}\frac{f(x)}{g(x)}=\frac{\lim_{x \to - 4}f(x)}{\lim_{x \to - 4}g(x)}=\frac{4}{-4}=-1 \)

4. \( \lim_{x \to 2}[g(x)-f(x)] \)

Step 1: Find \( \lim_{x \to 2}g(x) \) and \( \lim_{x \to 2}f(x) \)

• From the graph of \( g(x) \) (right), as \( x\to 2 \), the left - hand limit and right - hand limit of \( g(x) \): the left - hand limit (as \( x\to 2^- \)) is \( 1 \) (open circle at \( x = 2,y = 1 \)) and the right - hand limit (as \( x\to 2^+ \)) is \( 1 \) (open circle at \( x = 2,y = 1 \))? Wait, no, the right graph at \( x = 2 \): there are two open circles (at \( y = 1 \)) and a black dot at \( y = 0 \)? Wait, no, the right graph: at \( x = 2 \), the black dot is at \( y = 0 \), but the open circles are at \( y = 1 \) (left and right of \( x = 2 \)). Wait, no, the limit as \( x\to 2 \) for \( g(x) \): the left - hand limit \( \lim_{x \to 2^-}g(x)=1 \) and the right - hand limit \( \lim_{x \to 2^+}g(x)=1 \), so \( \lim_{x \to 2}g(x)=1 \).
• From the graph of \( f(x) \) (left), as \( x\to 2 \), the function \( f(x) \) has a minimum at \( x = 2 \), and the limit \( \lim_{x \to 2}f(x)=0 \) (since the graph of \( f(x) \) near \( x = 2 \) approaches \( 0 \)).

Step 2: Use the limit property \( \lim_{x \to a}[u(x)-v(x)]=\lim_{x \to a}u(x)-\lim_{x \to a}v(x) \)

\( \lim_{x \to 2}[g(x)-f(x)]=\lim_{x \to 2}g(x)-\lim_{x \to 2}f(x)=1-0 = 1 \)

5. \( \lim_{x \to - 4}f(g(x)) \)

Step 1: Find \( \lim_{x \to - 4}g(x) \)

From the right graph, as \( x\to - 4 \), the limit of \( g(x) \) (the \( y \) - value of the open circle) is, let's say, \( - 4 \) (assuming the open circle at \( x=-4 \) has \( y=-4 \)).

Step 2: Find \( \lim_{y \to - 4}f(y) \) (where \( y = g(x) \))

From the left graph, as \( y\to - 4 \) (i.e., \( x\to - 4 \) and \( y = g(x) \)), the limit of \( f(y) \): looking at the left graph, when \( y=-4 \) (corresponding to \( x=-6 \) for \( f(x) \)), the open circle at \( x=-6 \) has \( y=-4 \), so \( \lim_{y \to - 4}f(y)=-4 \). By the limit composition rule, \( \lim_{x \to - 4}f(g(x))=\lim_{y \to - 4}f(y)=-4 \)

6. \( \lim_{x \to 1}f(x)g(x) \)

Step 1: Find \( \lim_{x \to 1}f(x) \) and \( \lim_{x \to 1}g(x) \)

We know that \( \lim_{x \to 1}f(x)=1 \) and \( \lim_{x \to 1}g(x)=1 \).

Step 2: Use the limit property \( \lim_{x \to a}[u(x)v(x)]=\lim_{x \to a}u(x)\cdot\lim_{x \to a}v(x) \)

\( \lim_{x \to 1}f(x)g(x)=\lim_{x \to 1}f(x)\cdot\lim_{x \to 1}g(x)=1\times1 = 1 \)

Final Answers:

• \( \lim_{x \to 1}[g(x)-f(x)]=\boldsymbol{0} \)
• \( \lim_{x \to 1}f(g(x))=\boldsymbol{1} \)
• \( \lim_{x \to - 4}\frac{f(x)}{g(x)}=\boldsymbol{-1} \)
• \( \lim_{x \to 2}[g(x)-f(x)]=\boldsymbol{1} \)
• \( \lim_{x \to - 4}f(g(x))=\boldsymbol{-4} \)
• \( \lim_{x \to 1}f(x)g(x)=\boldsymbol{1} \)