Chapter 12

The value of the function, the output, at $x=a$ is $f\left(a\right).$ When the $\underset{x\to a}{\lim }f\left(x\right)$ 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.

–4

–4

2

does not exist

4

does not exist

Answers will vary.

Answers will vary.

Answers will vary.

Answers will vary.

Answers will vary.

Answers will vary.

Answers will vary.

7.38906

54.59815

$e^6\approx 403.428794,$ $e^7\approx 1096.633158,$ $e^n$

$\underset{x\to -2}{\lim }f(x)=1$

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

$\underset{x\to 1}{\lim }\left(\frac{x^2-1}{x^2-3x+2}\right)=-2.00$

$\underset{x\to 1}{\lim }\left(\frac{10-10x^2}{x^2-3x+2}\right)=20.00$

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

$\underset{x\to 0}{\lim }\frac{7\tan x}{3x}=\frac{7}{3}$

$\underset{x\to 0}{\lim }{e}^{e^{-\frac{1}{x^2}}}=1.0$

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

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

$\underset{x\to 0}{\lim }\frac{5}{1-e^{\frac{2}{x}}}$ 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,

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.

$\frac{-10}{3}$

6

$\frac{1}{2}$

6

does not exist

$-12$

$-\frac{\sqrt{5}}{10}$

$-108$

1

6

1

1

does not exist

$6+\sqrt{5}$

$\frac{3}{5}$

0

$-3$

does not exist; right-hand limit is not the same as the left-hand limit.

2

Limit does not exist; limit approaches infinity.

$4x+2h$

$2x+h+4$

$\frac{\cos (x+h)-\cos (x)}{h}$

$\frac{-1}{x(x+h)}$

$\frac{-1}{\sqrt{x+h}+\sqrt{x}}$

$f\left(x\right)=\frac{x^2+5x+6}{x+3}$

does not exist

52

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 $a=-3$; $f(-3)$ does not exist

removable discontinuity at $a=-4$; $f(-4)$ is not defined

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

$\underset{x\to 2}{\lim }f(x)$ does not exist.

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

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

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

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

$f\left(2\right)$ is not defined.

$f\left(-3\right)$ is not defined.

$f\left(0\right)$ is not defined.

Continuous on $(-\infty ,\infty )$

Continuous on $(-\infty ,\infty )$

Discontinuous at $x=0$ and $x=2$

Discontinuous at $x=0$

Continuous on $(0,\infty )$

Continuous on $[4,\infty )$

Continuous on $(-\infty ,\infty )$.

1, but not 2 or 3

1 and 2, but not 3

$f\left(0\right)$ is undefined.

$(-\infty ,0)\cup (0,\infty )$

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.

$\frac{x^3+6x^2-7x}{\left(x+7\right)\left(x-1\right)}$

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

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 $f(a)+45(b-a).$

$f^{\prime }(x)=-2$

$f^{\prime }(x)=4x+1$

$f^{\prime }(x)=\frac{1}{{(x-2)}^2}$

$\frac{-16}{{\left(3+2x\right)}^2}$

$f^{\prime }(x)=9x^2-2x+2$

$f^{\prime }(x)=0$

$-\frac{1}{3}$

undefined

$f^{\prime }(x)=-6x-7$

$f^{\prime }(x)=9x^2+4x+1$

$y=12x-15$

$k=-10$ or $k=2$

Discontinuous at $x=-2$ and $x=0.$ Not differentiable at –2, 0, 2.

Discontinuous at $x=5.$ Not differentiable at -4, –2, 0, 1, 3, 4, 5.

$f\left(0\right)=-2$

$f\left(2\right)=-6$

$f^{\prime }\left(-1\right)=9$

$f^{\prime }\left(1\right)=-3$

$f^{\prime }\left(3\right)=9$

Answers vary. The slope of the tangent line near $x=1$ is 2.

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 after 2 seconds is 96 feet.

The height of the projectile at $t=3$ seconds is 96 feet.

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.

$36\pi$

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

$21 per unit

$36

$f\text{'}(x)=10a-1$

$\frac{4}{{\left(3-x\right)}^2}$

2

does not exist

Discontinuous at $x=-1(\underset{x\to a}{\lim }f(x)$ does not exist $),x=3($jump discontinuity$)$, and$x=7(\underset{x\to a}{\lim }f(x)$ does not exist$).$

$\begin{array}{ll}\underset{x\to -2}{\lim }\hfill & f(x)=0\hfill \end{array}$

Does not exist

$-\frac{3}{5}$

$1$

$6$

$500$

$-\frac{6}{7}$

At $x=4,$ the function has a vertical asymptote.

At $x=3,$ the function has a vertical asymptote.

Removable discontinuity at $a=9$

Removable discontinuity at $x=5$

Removable discontinuity at $x=5$, discontinuity at $x=1$

Removable discontinuity at $x=-2$, discontinuity at $x=5$

$3$

$\frac{1}{(x+1)(x+h+1)}$

$\frac{e^{2x+2h}-e^{2x}}{h}$

$10x-3$

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

3

0

$\mathrm{-1}$

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

$-\frac{1}{50}$

$1$

Removable discontinuity at $x=3$

$f\text{'}\left(x\right)=-\frac{3}{2a\frac{3}{2}}$

Discontinuous at −2, 0, not differentiable at −2, 0, 2

Not differentiable at $x=0$ (no limit)

The height of the projectile at $t=2$ Seconds

The average velocity from $t=1$ $t=2$

$\frac{1}{3}$

$0$

2

$x=1$

$y=-14x-18$

The graph is not differentiable at $x=1$ (cusp).

$f^{\text{'}}(x)=8x$

$f^{\text{'}}(x)=-\frac{1}{{\left(2+x\right)}^2}$

$f^{\text{'}}(x)=-3x^2$

$f\text{'}(x)=\frac{1}{2\sqrt{x-1}}$