함수의 극한과 연속 전체 문제

The velocity is positive when the object is going in the increasing direction; it is going “up" on the graph when $t$ is between 0 and 2, and when $t$ is between $6$ and $7$. So, the velocity is positive when $t$ is in $(0,2) \cup (6,7)$.

(a) Average velocity: $\dfrac{\text{change in position}}{\text{change in time}} = \dfrac{s(5)-s(3)}{5-3} = \dfrac{(3\cdot 5^2+5)-(3\cdot 3^2+5)}{5-3}=24$ units per second. (b) From the notes, we know the velocity of an object at time $a$ is $$v(a)=\lim_{h \rightarrow 0}\frac{s(a+h)-s(a)}{h}$$ So, in our case: $$v(1)=\lim_{h \rightarrow 0}\frac{s(1+h)-s(1)}{h} = \lim_{h \rightarrow 0}\frac{[3(1+h)^2+5]-[3(1)^2+5]}{h} = \lim_{h \rightarrow 0}\frac{6h+3h^2}{h}= \lim_{h \rightarrow 0}6+3h = 6$$ So the velocity when $t=1$ is 6 units per second.

(a) Average velocity: $\dfrac{\text{change in position}}{\text{change in time}} = \dfrac{s(9)-s(1)}{9-1} = \dfrac{3-1}{9-1}=\dfrac{1}{4}$ units per second. (b) From the notes, we know the velocity of an object at time $a$ is $$v(a)=\lim_{h \rightarrow 0}\frac{s(a+h)-s(a)}{h}$$ So, in our case:

$$\begin{aligned} v(1)=\lim_{h \rightarrow 0}\frac{s(1+h)-s(1)}{h} &=\lim_{h \rightarrow 0}\frac{\sqrt{1+h}-1}{h}\\ &=\lim_{h \rightarrow 0}\frac{\sqrt{1+h}-1}{h}\cdot \left( \dfrac{\sqrt{1+h}+1}{\sqrt{1+h}+1}\right)\\ &=\lim_{h \rightarrow 0}\frac{(1+h)-1}{h(\sqrt{1+h}+1)}\\ &=\lim_{h \rightarrow 0}\frac{1}{\sqrt{1+h}+1}=\frac{1}{2}\text{ units per second} \end{aligned}$$ (c) $$\begin{aligned} v(9)=\lim_{h \rightarrow 0}\frac{s(9+h)-s(9)}{h} &=\lim_{h \rightarrow 0}\frac{\sqrt{9+h}-3}{h}\\ &=\lim_{h \rightarrow 0}\frac{\sqrt{9+h}-3}{h}\cdot \left( \dfrac{\sqrt{9+h}+3}{\sqrt{9+h}+3}\right)\\ &=\lim_{h \rightarrow 0}\frac{(9+h)-9}{h(\sqrt{9+h}+3)}\\ &=\lim_{h \rightarrow 0}\frac{1}{\sqrt{9+h}+3}=\frac{1}{6}\text{ units per second} \end{aligned}$$

Remark: the average velocity is not the average of the two instantaneous velocities.

(a) $\displaystyle \lim_{x \rightarrow -2} f(x)=1$: as $x$ gets very close to $-2$, $y$ gets very close to $1$. (b) $\displaystyle \lim_{x \rightarrow 0}f(x)=0$: as $x$ gets very close to $0$, $y$ also gets very close to $0$. (c) $\displaystyle \lim_{x \rightarrow 2}f(x)=2$: as $x$ gets very close to $2$, $y$ gets very close to $2$. We ignore the value of the function where $x$ is exactly $2$.

The limit does not exist. As $x$ approaches 0 from the left, $y$ approaches -1; as $x$ approaches 0 from the right, $y$ approaches 1. This tells us $\displaystyle\lim_{x \rightarrow 0^-} f(x)=-1$ and $\displaystyle\lim_{x \rightarrow 0^+} f(x)=1$, but neither of these are what the question asked. Since the limits from left and right do not agree, the limit does not exist. Put another way, there is no single number $y$ approaches as $x$ approaches 0, so the limit $\displaystyle\lim_{x \rightarrow 0} f(x)$ does not exist.

(a) $\displaystyle \lim_{x \rightarrow -1^{-}} f(x)=2$: as $x$ approaches $-1$ from the left, $y$ approaches 2. It doesn't matter that the function isn't defined at $x=-1$, and it doesn't matter what happens to the right of $x=-1$. (b) $\displaystyle \lim_{x \rightarrow -1^{+}} f(x)=-2$: as $x$ approaches $-1$ from the right, $y$ approaches -2. It doesn't matter that the function isn't defined at $-1$, and it doesn't matter what happens to the left of $-1$. (c) $\displaystyle \lim_{x \rightarrow -1} f(x) =$ DNE: since the limits from the left and right don't agree, the limit does not exist. (d) $\displaystyle \lim_{x \rightarrow -2^{+}} f(x) =0$: as $x$ approaches $-2$ from the right, $y$ approaches 0. It doesn't matter that the function isn't defined at 2, or to the left of 2. (e) $\displaystyle \lim_{x \rightarrow 2^{-}} f(x)=0$: as $x$ approaches $2$ from the left, $y$ approaches 0. It doesn't matter that the function isn't defined at 2, or to the right of 2.

Many answers are possible; here is one.

As $x$ gets closer and closer to 3, $y$ gets closer and closer to 10: this shows $\displaystyle\lim_{x \rightarrow 3} f(x)=10$. Also, at 3 itself, the function takes the value 10; this shows $f(3)=10$.

Many answers are possible; here is one.

Note that, as $x$ gets closer and closer to 3 except at 3 itself, $y$ gets closer and closer to 10: this shows $\displaystyle\lim_{x \rightarrow 3} f(x)=10$. Then, when $x=3$, the function has value 0: this shows $f(3)=0$.

$\displaystyle\lim_{x \rightarrow -2^-} f(x)=16$: in order for the limit $\displaystyle\lim_{x \rightarrow 2} f(x)$ to exist and be equal to 16, both one sided limits must exist and be equal to 16.

Not enough information to say. If $\displaystyle\lim_{x \rightarrow -2^+} f(x)=16$, then $\displaystyle\lim_{x \rightarrow -2} f(x)=16$. If $\displaystyle\lim_{x \rightarrow -2^+} f(x)\neq 16$, then $\displaystyle\lim_{x \rightarrow -2} f(x)$ does not exist.

$\displaystyle\lim_{t \rightarrow 0} \sin t=0$: as $t$ approaches 0, $\sin t$ approaches 0 as well.

$\displaystyle\lim_{x \rightarrow 0^+} \log x = -\infty$: as $x$ approaches 0 from the right, $\log x$ is negative and increasingly large, growing without bound.

$\displaystyle\lim_{y \rightarrow 3} y^2=9$: as $y$ gets closer and closer to 3, $y^2$ gets closer and closer to $3^2$.

$\displaystyle\lim_{x \rightarrow 0^-} \dfrac{1}{x}=-\infty$: as $x$ gets closer and closer to 0 from the left, $\dfrac{1}{x}$ becomes a larger and larger negative number.

$\displaystyle\lim_{x \rightarrow 3} \dfrac{1}{10}=\dfrac{1}{10}$: no matter what $x$ is, $\dfrac{1}{10}$ is always $\dfrac{1}{10}$. In particular, as $x$ approaches 3, $\dfrac{1}{10}$ stays put at $\dfrac{1}{10}$.

When $x$ is very close to 3, $f(x)$ looks like the function $x^2$. So: $\displaystyle\lim_{x \rightarrow 3} f(x) = \displaystyle\lim_{x \rightarrow 3} x^2=9$

The statement $\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=10$ tells us that, as $x$ gets very close to 3, $f(x)$ is 10 times as large as $g(x)$. We notice that if $f(x)=10g(x)$, then $\dfrac{f(x)}{g(x)}=10$, so $\displaystyle\lim_{x \rightarrow} \dfrac{f(x)}{g(x)}=10$ wherever $f$ and $g$ exist. So it's enough to find a function $g(x)$ that has limit 0 at 3. Such a function is (for example) $g(x)=x-3$. So, we take $f(x)=10(x-3)$ and $g(x)=x-3$. It is easy now to check that $\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$ and $\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=\displaystyle \lim_{x \to 3}\frac{10(x-3)}{x-3}=\displaystyle \lim_{x \to 3}10=10$.

One way to start this problem is to remember $\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x^2}=\infty$. (Using $\dfrac{1}{x^2}$ as opposed to $\dfrac{1}{x}$ is important, since $\displaystyle\lim_{x \rightarrow 0}\dfrac{1}{x}$ does not exist.) Then by “shifting" by three, we find $\displaystyle \lim_{x \rightarrow 3} \dfrac{1}{(x-3)^2}=\infty$. So it is enough to arrange that $\dfrac{f(x)}{g(x)}=\dfrac{1}{(x-3)^2}$. We can achieve this with $f(x)=x-3$ and $g(x)=(x-3)^3$, and maintain $\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$. Again, this is one of many possible solutions.

We can begin by plotting the points that are easy to read off the diagram. $$\begin{array}{c|c|c} x&f(x)&\frac{1}{f(x)}\\ \hline -3&-3&\frac{-1}{3}\\ -2&0&UND\\ -1&3&\frac{1}{3}\\ 0&3&\frac{1}{3}\\ 1&\frac{3}{2}&\frac{2}{3}\\ 2&0&UND\\ 3&1&1 \end{array}$$ Note that $\frac{1}{f(x)}$ is undefined when $f(x) = 0$. So $\frac{1}{f(x)}$ is undefined at $x=-2$ and $x=2$. We shall look more closely at the behaviour of $\frac{1}{f(x)}$ for $x$ near $\pm 2$ shortly.

Plotting the above points, we get the following picture:

Since $f(x)$ is constant when $x$ is between -1 and 0, then also $\frac{1}{f(x)}$ is constant between -1 and 0, so we update our picture:

The big question that remains is the behaviour of $\frac{1}{f(x)}$ when $x$ is near -2 and 2. We can answer this question with limits. As $x$ approaches $-2$ from the left, $f(x)$ gets closer to zero, and is negative. So $\frac{1}{f(x)}$ will be negative, and will increase in magnitude without bound; that is, $\displaystyle\lim_{x \rightarrow -2^-}\dfrac{1}{f(x)}=-\infty$. Similarly, as $x$ approaches $-2$ from the right, $f(x)$ gets closer to zero, and is positive. So $\frac{1}{f(x)}$ will be positive, and will increase in magnitude without bound; that is, $\displaystyle\lim_{x \rightarrow -2^+}\dfrac{1}{f(x)}=\infty$. We add this behaviour to our graph: