함수의 극한의 뜻

$$\begin{array}{cc}h & \frac{f(a+h)-f(a)}{h} \\ \hline -0.1 & 0.202027 \\ -0.01 & 0.2002 \\ 0.01 & 0.1998 \\ 0.1 & 0.198026\end{array}$$ The limit is approx. $0.2$.

$$\begin{array}{cc}h & \frac{f(a+h)-f(a)}{h} \\ \hline -0.1 & -0.998334 \\ -0.01 & -0.999983 \\ 0.01 & -0.999983 \\ 0.1 & -0.998334\end{array}$$ The limit is approx. $-1$.

$$\begin{array}{cc}h & \frac{f(a+h)-f(a)}{h} \\ \hline -0.1 & -0.0499583 \\ -0.01 & -0.00499996 \\ 0.01 & 0.00499996 \\ 0.1 & 0.0499583\end{array}$$ The limit is approx. $0.005$.

(a) True: since $y=2x+3$ is the tangent line to $y=f(x)$ at the point $x=2$, this means the function and the tangent line have the same value at $x=2$. So $f(2)=2(2)+3=7$. (b) In general, this is false. We are only guaranteed that the curve $y=f(x)$ and its tangent line $y=2x+3$ agree at $x=2$. The functions $f(x)$ and $2x+3$ may or may not take the same values when $x\ne 2$. For example, if $f(x)=2x+3$, then of course $f(x)$ and $2x+3$ agree for all values of $x$. But if $f(x) = 2x+3 +(x-2)^2$, then $f(x)$ and $2x+3$ agree only for $x=2$.

Since the tangent line to the curve at point $P$ passes through point $P$, the curve and the tangent line touch at point $P$. So, they must intersect at least once. By drawing various examples, we can see that different curves may touch their tangent lines exactly once, exactly twice, exactly three times, etc.

Speed is nonnegative; velocity has a sign (positive or negative) that indicates direction.

Yes–an object that is not moving has speed 0.

Since you started and ended in the same place, your difference in position was 0, and your difference in time was 24 hours. So, your average velocity was $\dfrac{0}{24}=0$ kph.

The slope of a curve is given by $\dfrac{\text{change in vertical component}}{\text{change in horizontal component}}$. The change in the vertical component is exactly $s(b)-s(a)$, and the change in the horizontal component is exactly $b-a$.

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.