Question

1. for the function $\phi$ graphed in the accompanying figure, find

(a) $\lim\limits_{x \to -2^-} \phi(x)$
(b) $\lim\limits_{x \to -2^+} \phi(x)$
(c) $\lim\limits_{x \to -2} \phi(x)$
(d) $\phi(-2)$
(e) $\lim\limits_{x \to -\infty} \phi(x)$
(f) $\lim\limits_{x \to +\infty} \phi(x)$.
figure ex-7

Explanation:

Part (a): $\boldsymbol{\lim_{x \to -2^-} \phi(x)}$

Step1: Analyze left - hand limit

As $x$ approaches $-2$ from the left (values less than $-2$), we look at the graph of $\phi(x)$ for $x < - 2$. From the graph, the left - hand side of $x=-2$ (the curve on the left of the vertical asymptote near $x = - 2$) approaches a value. By observing the horizontal behavior of the left - hand curve, we can see that as $x$ gets closer to $-2$ from the left, the function $\phi(x)$ approaches $1$ (assuming the horizontal line on the left has a $y$ - value of $1$).
$\lim_{x \to -2^-} \phi(x)=1$

Part (b): $\boldsymbol{\lim_{x \to -2^+} \phi(x)}$

Step1: Analyze right - hand limit

As $x$ approaches $-2$ from the right (values greater than $-2$), we look at the graph of $\phi(x)$ for $x > - 2$. The right - hand side of $x = - 2$ (the curve on the right of the vertical asymptote near $x=-2$) has a vertical asymptote? Wait, no, looking at the graph, when $x$ approaches $-2$ from the right, the function $\phi(x)$ goes to $+\infty$? Wait, no, maybe I misread. Wait, the graph has a vertical asymptote at $x=-2$? Wait, the left - hand curve (for $x < - 2$) is a horizontal - like curve approaching $y = 1$, and the right - hand curve (for $x>-2$) is a curve that starts from near the vertical asymptote (at $x = - 2$) and decreases towards $y = 0$ as $x\to+\infty$. Wait, maybe the vertical asymptote is at $x=-2$, so as $x\to - 2^+$, the function $\phi(x)$ approaches $+\infty$? Wait, no, the graph on the right of $x=-2$ (the part with $x > - 2$) has a curve that starts from a very high value (going to $+\infty$) as $x$ approaches $-2$ from the right? Wait, no, the original graph: the left - hand side (x < - 2) has a curve that is horizontal, maybe $y = 1$, and the right - hand side (x > - 2) has a curve that is a hyperbola - like curve, with a vertical asymptote at $x=-2$. So as $x\to - 2^+$, $\phi(x)\to+\infty$, and as $x\to - 2^-$, $\phi(x)\to1$? Wait, maybe I made a mistake. Wait, let's re - examine. The left - hand curve (x < - 2) is a horizontal line? Wait, the graph shows that for $x < - 2$, the function is a horizontal line (or a curve approaching a horizontal line) with $y = 1$. For $x > - 2$, the function is a curve that starts from near $x=-2$ (from the right) and decreases towards $y = 0$ as $x\to+\infty$, but as $x\to - 2^+$, the function goes to $+\infty$ (because it's a vertical asymptote at $x=-2$ for the right - hand side).
$\lim_{x \to -2^+} \phi(x)=+\infty$

Part (c): $\boldsymbol{\lim_{x \to -2} \phi(x)}$

Step1: Check left and right limits

For the limit as $x\to - 2$ to exist, the left - hand limit ($\lim_{x \to -2^-}\phi(x)$) and the right - hand limit ($\lim_{x \to -2^+}\phi(x)$) must be equal. We found that $\lim_{x \to -2^-}\phi(x) = 1$ and $\lim_{x \to -2^+}\phi(x)=+\infty$. Since $1
eq+\infty$, the limit $\lim_{x \to -2}\phi(x)$ does not exist.
$\lim_{x \to -2} \phi(x)$ does not exist

Part (d): $\boldsymbol{\phi(-2)}$

Answer:

s:
(a) $\boldsymbol{1}$

(b) $\boldsymbol{+\infty}$

(c) $\boldsymbol{\text{Does not exist}}$

(d) $\boldsymbol{\text{Undefined}}$

(e) $\boldsymbol{1}$

(f) $\boldsymbol{0}$