5. use the graph of $f(x)$ below to calculate each limit. you do not need to show your work. answer all parts (a) - (g). if a limit doesn’t exist, write \dne\. (a) $limlimits_{x o -3^{-}} f(x) =$ (b) $limlimits_{x o -3^{+}} f(x) =$ (c) $limlimits_{x o -3} f(x) =$ (d) $limlimits_{x o 0^{-}} f(x) =$ (e) $limlimits_{x o 0^{+}} f(x) =$ (f) $limlimits_{x o 0} f(x) =$ (g) $limlimits_{x o infty} f(x) =$

Question

5. use the graph of $f(x)$ below to calculate each limit. you do not need to show your work. answer all parts (a) - (g). if a limit doesn’t exist, write \dne\. (a) $limlimits_{x	o -3^{-}} f(x) =$ (b) $limlimits_{x	o -3^{+}} f(x) =$ (c) $limlimits_{x	o -3} f(x) =$ (d) $limlimits_{x	o 0^{-}} f(x) =$ (e) $limlimits_{x	o 0^{+}} f(x) =$ (f) $limlimits_{x	o 0} f(x) =$ (g) $limlimits_{x	o infty} f(x) =$

Accepted Answer

### Part (a) # Explanation: ## Step1: Analyze left - hand limit as $x\to - 3^{-}$ The left - hand limit as $x$ approaches $-3$ from the left (values less than $-3$) is determined by the graph of the function for $x < - 3$. The graph on the left side of $x=-3$ (the upper part of the graph) has a parabola - like shape. As $x$ approaches $-3$ from the left, we look at the $y$ - value that the function approaches. From the graph, as $x\to - 3^{-}$, the function values approach $+\infty$ (since the graph is going upwards as $x$ approaches $-3$ from the left). But wait, no, actually, looking at the vertical asymptote at $x = - 3$, the left - hand side (the upper curve) as $x\to - 3^{-}$ is going to $+\infty$? Wait, no, the upper curve: when $x$ approaches $-3$ from the left (values like $-4,-5,\cdots$ moving towards $-3$), the upper curve (the one with the two arrows going up) at $x=-6$ has a hole, but as $x$ moves towards $-3$ from the left along the upper curve, the $y$ - value is increasing? Wait, no, the vertical asymptote is at $x=-3$. The left - hand limit (from $x < - 3$): the upper graph (the one with the two upward - pointing arrows) as $x\to - 3^{-}$ is approaching $+\infty$? Wait, no, the lower graph (the one with the downward - pointing arrow at $x=-3$ from the right) is going to $-\infty$ as $x\to - 3^{+}$, but for $x\to - 3^{-}$, the upper graph (the one with the two upward arrows) is approaching $+\infty$? Wait, no, let's re - examine. The upper graph: when $x$ is less than $-3$ (left of $x = - 3$), the graph is a parabola - like curve opening upwards. As $x$ approaches $-3$ from the left ($x\to - 3^{-}$), the $y$ - values of this curve go to $+\infty$ (since the curve is moving upwards as $x$ gets closer to $-3$ from the left). But wait, the vertical asymptote at $x=-3$: the left - hand limit (from $x < - 3$) of the upper curve is $+\infty$? Wait, no, maybe I made a mistake. Wait, the upper curve: when $x$ is in $(-\infty,-3)$, the graph is a parabola with vertex around $x=-6$. As $x$ approaches $-3$ from the left, the distance from $x=-6$ to $x=-3$ is $3$ units. The parabola $y=a(x + 6)^{2}+k$ (vertex at $(-6,1)$ maybe? The hole is at $x=-6$, $y = 1$ (the open circle) and the dot is above it. Wait, no, the open circle at $x=-6$ has $y = 1$ (the lower part of the open circle) and the dot is above it ( $y = 3$ maybe). But as $x$ approaches $-3$ from the left (along $x < - 3$), the upper curve (the one with the two upward arrows) is moving towards $+\infty$ because it's a parabola opening upwards with vertex at $x=-6$, so as $x$ moves away from the vertex towards $-3$ (which is to the right of the vertex $x=-6$), the $y$ - value increases. So $\lim_{x\to - 3^{-}}f(x)=\infty$? But wait, the vertical asymptote at $x=-3$, the left - hand limit (from $x < - 3$) is $+\infty$. Wait, no, maybe I messed up. Let's start over. For part (a), $\lim_{x\to - 3^{-}}f(x)$: we look at the graph for $x$ values less than $-3$ (approaching $-3$ from the left). The graph on the left side of $x=-3$ (the upper graph) is a curve that as $x$ gets closer to $-3$ from the left, the $y$ - values are going to $+\infty$ (since the graph is rising as $x$ approaches $-3$ from the left). So $\lim_{x\to - 3^{-}}f(x)=\infty$? But maybe I was wrong. Wait, the vertical asymptote at $x=-3$: the left - hand side ( $x < - 3$) of the asymptote, the function (the upper curve) is going to $+\infty$ as $x\to - 3^{-}$, and the right - hand side ( $x > - 3$) of the asymptote, the function (the lower curve) is going to $-\infty$ as $x\to - 3^{+}$. ## Step2: Conclusion for part (a) $\lim_{x\to - 3^{-}}f(x)=\infty$ (or we can say it does not exist in the real - number sense, but in the context of limits, we can say it approaches $+\infty$). But wait, maybe the graph is different. Wait, the upper graph: when $x=-6$, there is a hole, and the dot is above it. As $x$ moves from $-6$ towards $-3$ (left to right along the upper curve), the $y$ - value increases. So as $x\to - 3^{-}$, the function values go to $+\infty$. ### Part (b) # Explanation: ## Step1: Analyze right - hand limit as $x\to - 3^{+}$ The right - hand limit as $x$ approaches $-3$ from the right (values greater than $-3$) is determined by the graph of the function for $x > - 3$. The graph on the right side of $x=-3$ (the lower part of the graph) has a curve that is going downwards as $x$ approaches $-3$ from the right. As $x$ approaches $-3$ from the right (values like $-2,-1,0,\cdots$ moving towards $-3$), the $y$ - value that the function approaches is $-\infty$ (since the graph is going downwards as $x$ approaches $-3$ from the right). ## Step2: Conclusion for part (b) $\lim_{x\to - 3^{+}}f(x)=-\infty$ ### Part (c) # Explanation: ## Step1: Recall the definition of the limit at a point For the limit $\lim_{x\to a}f(x)$ to exist, $\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)$. We found that $\lim_{x\to - 3^{-}}f(x)=\infty$ and $\lim_{x\to - 3^{+}}f(x)=-\infty$. Since $\lim_{x\to - 3^{-}}f(x) eq\lim_{x\to - 3^{+}}f(x)$ (one is $+\infty$ and the other is $-\infty$), the two - sided limit $\lim_{x\to - 3}f(x)$ does not exist. ## Step2: Conclusion for part (c) $\lim_{x\to - 3}f(x)=\text{DNE}$ ### Part (d) # Explanation: ## Step1: Analyze left - hand limit as $x\to 0^{-}$ The left - hand limit as $x$ approaches $0$ from the left (values less than $0$) is determined by the graph of the function for $x < 0$. The graph for $x < 0$ (the curve that approaches the open circle at $x = 0$ from the left) has a $y$ - value of $-3$ (the open circle at $x = 0$ from the left has a $y$ - coordinate of $-3$). As $x$ approaches $0$ from the left ($x\to 0^{-}$), the function values approach $-3$. ## Step2: Conclusion for part (d) $\lim_{x\to 0^{-}}f(x)=-3$ ### Part (e) # Explanation: ## Step1: Analyze right - hand limit as $x\to 0^{+}$ The right - hand limit as $x$ approaches $0$ from the right (values greater than $0$) is determined by the graph of the function for $x > 0$. The graph for $x > 0$ (the line segment that starts at the point $(0,-5)$ and goes to the right) as $x$ approaches $0$ from the right, the function values approach $-5$ (since the point at $x = 0$ on the right - hand side is $(0,-5)$ (the closed dot)). Wait, no: the open circle at $x = 0$ from the left is $-3$, and the closed dot at $x = 0$ is $-5$. As $x$ approaches $0$ from the right ($x\to 0^{+}$), we look at the graph for $x>0$. The graph for $x > 0$ starts at $x = 0$ with $y=-5$ (the closed dot) and goes up. So as $x\to 0^{+}$, the function values approach $-5$. ## Step2: Conclusion for part (e) $\lim_{x\to 0^{+}}f(x)=-5$ ### Part (f) # Explanation: ## Step1: Recall the definition of the two - sided limit For the limit $\lim_{x\to 0}f(x)$ to exist, $\lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{+}}f(x)$. We found that $\lim_{x\to 0^{-}}f(x)=-3$ and $\lim_{x\to 0^{+}}f(x)=-5$. Since $-3 eq - 5$, the two - sided limit $\lim_{x\to 0}f(x)$ does not exist. ## Step2: Conclusion for part (f) $\lim_{x\to 0}f(x)=\text{DNE}$ ### Part (g) # Explanation: ## Step1: Analyze the limit as $x\to\infty$ The limit as $x$ approaches $\infty$ ( $x$ becomes very large positive) is determined by the graph of the function for large positive $x$. The graph for large positive $x$ (the curve that has an open circle and then goes downwards) has a horizontal asymptote? Wait, the graph for $x$ large positive is a curve that is approaching $-7$ (the dashed line at $y=-7$)? Wait, no, the dashed line is at $y = - 7$, but the curve for $x\to\infty$ is going towards $-7$? Wait, the graph on the right - hand side (for large $x$) has a curve that is approaching $y=-7$ as $x\to\infty$? Wait, no, the arrow at the end of the right - hand curve is going towards $-7$? Wait, the $y$ - axis has a dashed line at $y=-7$. The right - hand curve (the one with the open circle at some $x$ and then going to the right) as $x$ becomes very large, the $y$ - value approaches $-7$. ## Step2: Conclusion for part (g) $\lim_{x\to\infty}f(x)=-7$ # Answers: (a) $\lim_{x\to - 3^{-}}f(x)=\infty$ (or if we consider the context, maybe the graph is different, but based on the vertical asymptote, the left - hand limit as $x\to - 3^{-}$ is $+\infty$) (b) $\lim_{x\to - 3^{+}}f(x)=-\infty$ (c) $\lim_{x\to - 3}f(x)=\text{DNE}$ (d) $\lim_{x\to 0^{-}}f(x)=-3$ (e) $\lim_{x\to 0^{+}}f(x)=-5$ (f) $\lim_{x\to 0}f(x)=\text{DNE}$ (g) $\lim_{x\to\infty}f(x)=-7$ Wait, there are some possible mistakes in the analysis of part (a) and (b). Let's re - check: For part (a): The upper graph (left of $x=-3$): when $x$ approaches $-3$ from the left, the graph is going up, so $\lim_{x\to - 3^{-}}f(x)=\infty$ (correct). For part (b): The lower graph (right of $x=-3$): when $x$ approaches $-3$ from the right, the graph is going down, so $\lim_{x\to - 3^{+}}f(x)=-\infty$ (correct). For part (d): The left - hand limit as $x\to 0^{-}$: the graph from the left of $x = 0$ ( $x<0$) approaches the open circle at $x = 0$ with $y=-3$, so $\lim_{x\to 0^{-}}f(x)=-3$ (correct). For part (e): The right - hand limit as $x\to 0^{+}$: the graph from the right of $x = 0$ ( $x > 0$) starts at the closed dot $(0,-5)$, so $\lim_{x\to 0^{+}}f(x)=-5$ (correct). For part (g): The right - hand curve as $x\to\infty$ is approaching $y=-7$ (the dashed line), so $\lim_{x\to\infty}f(x)=-7$ (correct).

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Analyze left - hand limit as \(x\to - 3^{-}\)

Step1: Analyze right - hand limit as \(x\to - 3^{+}\)

Step2: Conclusion for part (b)

Part (c)

Step1: Recall the definition of the limit at a point

Step2: Conclusion for part (c)

Part (d)

Answer: