### Try It

### 12.1 Finding Limits: Numerical and Graphical Approaches

### 12.2 Finding Limits: Properties of Limits

### 12.3 Continuity

### 12.4 Derivatives

- After zero seconds, she has traveled 0 feet.
- After 10 seconds, she has traveled 150 feet east.
- After 10 seconds, she is moving eastward at a rate of 15 ft/sec.
- After 20 seconds, she is moving westward at a rate of 10 ft/sec.
- After 40 seconds, she is 100 feet westward of her starting point.

The graph of $f$ is continuous on $\left(-\infty ,1\right)\cup \left(1,3\right)\cup \left(3,\infty \right).$ The graph of $f$ is discontinuous at $x=1$ and $x=3.$ The graph of $f$ is differentiable on $\left(-\infty ,1\right)\cup \left(1,3\right)\cup \left(3,\infty \right).$ The graph of $f$ is not differentiable at $x=1$ and $x=3.$

### 12.1 Section Exercises

The value of the function, the output, at $x=a$ is $f\left(a\right).$ When the $\underset{x\to a}{\mathrm{lim}}f\left(x\right)$ is taken, the values of $x$ get infinitely close to $a$ but never equal $a.$ As the values of $x$ approach $a$ from the left and right, the limit is the value that the function is approaching.

$\underset{x\to 3}{\mathrm{lim}}\left(\frac{{x}^{2}-x-6}{{x}^{2}-9}\right)=\frac{5}{6}\approx 0.83$

$\underset{x\to \frac{-1}{2}}{\mathrm{lim}}\left(\frac{x}{4{x}^{2}+4x+1}\right)$ does not exist. Function values decrease without bound as $x$ approaches –0.5 from either left or right.

$\underset{x\to -{1}^{-}}{\mathrm{lim}}\frac{\left|x+1\right|}{x+1}=\frac{-(x+1)}{(x+1)}=-1$ and $\underset{x\to -{1}^{+}}{\mathrm{lim}}\frac{\left|x+1\right|}{x+1}=\frac{(x+1)}{(x+1)}=1;$ since the right-hand limit does not equal the left-hand limit, $\underset{x\to -1}{\mathrm{lim}}\frac{\left|x+1\right|}{x+1}$ does not exist.

$\underset{x\to -1}{\mathrm{lim}}\frac{1}{{\left(x+1\right)}^{2}}$ does not exist. The function increases without bound as $x$ approaches $-1$ from either side.

$\underset{x\to 0}{\mathrm{lim}}\frac{5}{1-{e}^{\frac{2}{x}}}$ does not exist. Function values approach 5 from the left and approach 0 from the right.

Through examination of the postulates and an understanding of relativistic physics, as $v\to c,$ $m\to \infty .$ Take this one step further to the solution,

### 12.2 Section Exercises

If $f$ is a polynomial function, the limit of a polynomial function as $x$ approaches $a$ will always be $f\left(a\right).$

It could mean either (1) the values of the function increase or decrease without bound as $x$ approaches $c,$ or (2) the left and right-hand limits are not equal.

### 12.3 Section Exercises

Informally, if a function is continuous at $x=c,$ then there is no break in the graph of the function at $f\left(c\right),$ and $f\left(c\right)$ is defined.

Discontinuous at $a=3$; $\underset{x\to 3}{\mathrm{lim}}f(x)=3,$ but $f(3)=6,$ which is not equal to the limit.

$\underset{x\to {1}^{-}}{\mathrm{lim}}f(x)=4;\underset{x\to {1}^{+}}{\mathrm{lim}}f(x)=1$. Therefore, $\underset{x\to 1}{\mathrm{lim}}f(x)$ does not exist.

$\underset{x\to {1}^{-}}{\mathrm{lim}}f(x)=5\ne \underset{x\to {1}^{+}}{\mathrm{lim}}f(x)=-1$. Thus $\underset{x\to 1}{\mathrm{lim}}f(x)$ does not exist.

$\underset{x\to -{3}^{-}}{\mathrm{lim}}f(x)=-6$, $\underset{x\to -{3}^{+}}{\mathrm{lim}}f(x)=-\frac{1}{3}$

Therefore, $\underset{x\to -3}{\mathrm{lim}}f(x)$ does not exist.

At $x=-1,$ the limit does not exist. At $x=1,$ $f\left(1\right)$ does not exist.

At $x=2,$ there appears to be a vertical asymptote, and the limit does not exist.

The function is discontinuous at $x=1$ because the limit as $x$ approaches 1 is 5 and $f\left(1\right)=2.$

### 12.4 Section Exercises

The slope of a linear function stays the same. The derivative of a general function varies according to $x.$ Both the slope of a line and the derivative at a point measure the rate of change of the function.

Average velocity is 55 miles per hour. The instantaneous velocity at 2:30 p.m. is 62 miles per hour. The instantaneous velocity measures the velocity of the car at an instant of time whereas the average velocity gives the velocity of the car over an interval.

The average rate of change of the amount of water in the tank is 45 gallons per minute. If $f\left(x\right)$ is the function giving the amount of water in the tank at any time $t,$ then the average rate of change of $f\left(x\right)$ between $t=a$ and $t=b$ is $f(a)+45(b-a).$

At 12:30 p.m., the rate of change of the number of gallons in the tank is –20 gallons per minute. That is, the tank is losing 20 gallons per minute.

At 200 minutes after noon, the volume of gallons in the tank is changing at the rate of 30 gallons per minute.

The height of the projectile is zero at $t=0$ and again at $t=5.$ In other words, the projectile starts on the ground and falls to earth again after 5 seconds.

$50.00 per unit, which is the instantaneous rate of change of revenue when exactly 10 units are sold.

### Review Exercises

$\begin{array}{l}\text{Discontinuousat}x=-1(\underset{x\to \phantom{\rule{0.8em}{0ex}}a\phantom{\rule{0.8em}{0ex}}}{\mathrm{lim}}f(x)\text{doesnotexist}),x=3\phantom{\rule{0.8em}{0ex}}(\text{jumpdiscontinuity}),\\ \text{and}x=7\phantom{\rule{0.8em}{0ex}}(\underset{x\to \phantom{\rule{0.8em}{0ex}}a\phantom{\rule{0.8em}{0ex}}}{\mathrm{lim}}f(x)\text{doesnotexist}).\end{array}$

The function would not be differentiable at however, 0 is not in its domain. So it is differentiable everywhere in its domain.

### Practice Test

$\underset{x\to {2}^{-}}{\mathrm{lim}}f(x)=-\frac{5}{2}a$ and $\underset{x\to {2}^{+}}{\mathrm{lim}}f(x)=9$ Thus, the limit of the function as $x$ approaches 2 does not exist.