Finding Limits: Numerical and Graphical Approaches

Intuitively, we know what a limit is. A car can go only so fast and no faster. A trash can might hold 33 gallons and no more. It is natural for measured amounts to have limits. What, for instance, is the limit to the height of a woman? The tallest woman on record was Jinlian Zeng from China, who was 8 ft 1 in.https://en.wikipedia.org/wiki/Human_height and http://en.wikipedia.org/wiki/List_of_tallest_people Is this the limit of the height to which women can grow? Perhaps not, but there is likely a limit that we might describe in inches if we were able to determine what it was.

To put it mathematically, the function whose input is a woman and whose output is a measured height in inches has a limit. In this section, we will examine numerical and graphical approaches to identifying limits.

Understanding Limit Notation

We have seen how a sequence can have a limit, a value that the sequence of terms moves toward as the number of terms increases. For example, the terms of the sequence

1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8} ...

gets closer and closer to 0. A sequence is one type of function, but functions that are not sequences can also have limits. We can describe the behavior of the function as the input values get close to a specific value. If the limit of a function $f (x) = L,$ then as the input $x$ gets closer and closer to $a,$ the output y-coordinate gets closer and closer to $L .$ We say that the output “approaches” $L .$

Figure 1 provides a visual representation of the mathematical concept of limit. As the input value $x$ approaches $a,$ the output value $f (x)$ approaches $L .$

Graph representing how a function with a hole at (a, L) approaches a limit. — Figure 1 The output (y--coordinate) approaches $L$ as the input (x-coordinate) approaches $a .$

We write the equation of a limit as

\lim_{x \to a} f (x) = L .

This notation indicates that as $x$ approaches $a$ both from the left of $x = a$ and the right of $x = a,$ the output value approaches $L .$

Consider the function

f (x) = \frac{x^{2} - 6 x - 7}{x - 7} .

We can factor the function as shown.

\begin{array}{l} f (x) = \frac{(x - 7) (x + 1)}{x - 7} & Cancel like factors in numerator and denominator. \\ f (x) = x + 1, x \neq 7 & Simplify. \end{array}

Notice that $x$ cannot be 7, or we would be dividing by 0, so 7 is not in the domain of the original function. In order to avoid changing the function when we simplify, we set the same condition, $x \neq 7,$ for the simplified function. We can represent the function graphically as shown in Figure 2.

Graph of an increasing function, f(x) = (x^2-6x-7)/(x-7), with a hole at (7, 8). — Figure 2 Because 7 is not allowed as an input, there is no point at $x = 7.$

What happens at $x = 7$ is completely different from what happens at points close to $x = 7$ on either side. The notation

\lim_{x \to 7} f (x) = 8

indicates that as the input $x$ approaches 7 from either the left or the right, the output approaches 8. The output can get as close to 8 as we like if the input is sufficiently near 7.

What happens at $x = 7 ?$ When $x = 7,$ there is no corresponding output. We write this as

f (7) does not exist.

This notation indicates that 7 is not in the domain of the function. We had already indicated this when we wrote the function as

f (x) = x + 1, x \neq 7.

Notice that the limit of a function can exist even when $f (x)$ is not defined at $x = a .$ Much of our subsequent work will be determining limits of functions as $x$ nears $a,$ even though the output at $x = a$ does not exist.

Example 1

Understanding the Limit of a Function

For the following limit, define $a, f (x),$ and $L .$

\lim_{x \to 2} (3 x + 5) = 11

Solution

First, we recognize the notation of a limit. If the limit exists, as $x$ approaches $a,$ we write

\lim_{x \to a} f (x) = L .

We are given

\lim_{x \to 2} (3 x + 5) = 11.

This means that $a = 2, f (x) = 3 x + 5, and L = 11.$

Analysis

Recall that $y = 3 x + 5$ is a line with no breaks. As the input values approach 2, the output values will get close to 11. This may be phrased with the equation $\lim_{x \to 2} (3 x + 5) = 11,$ which means that as $x$ nears 2 (but is not exactly 2), the output of the function $f (x) = 3 x + 5$ gets as close as we want to $3 (2) + 5,$ or 11, which is the limit $L,$ as we take values of $x$ sufficiently near 2 but not at $x = 2.$

Understanding Left-Hand Limits and Right-Hand Limits

We can approach the input of a function from either side of a value—from the left or the right. Figure 3 shows the values of

f (x) = x + 1, x \neq 7

as described earlier and depicted in Figure 2.

Table showing that f(x) approaches 8 from either side as x approaches 7 from either side. — Figure 3

Values described as “from the left” are less than the input value 7 and would therefore appear to the left of the value on a number line. The input values that approach 7 from the left in Figure 3 are $6.9,$ $6.99,$ and $6.999.$ The corresponding outputs are $7.9, 7.99,$ and $7.999.$ These values are getting closer to 8. The limit of values of $f (x)$ as $x$ approaches from the left is known as the left-hand limit. For this function, 8 is the left-hand limit of the function $f (x) = x + 1, x \neq 7$ as $x$ approaches 7.

Values described as “from the right” are greater than the input value 7 and would therefore appear to the right of the value on a number line. The input values that approach 7 from the right in Figure 3 are $7.1,$ $7.01,$ and $7.001.$ The corresponding outputs are $8.1,$ $8.01,$ and $8.001.$ These values are getting closer to 8. The limit of values of $f (x)$ as $x$ approaches from the right is known as the right-hand limit. For this function, 8 is also the right-hand limit of the function $f (x) = x + 1, x \neq 7$ as $x$ approaches 7.

Figure 3 shows that we can get the output of the function within a distance of 0.1 from 8 by using an input within a distance of 0.1 from 7. In other words, we need an input $x$ within the interval $6.9 < x < 7.1$ to produce an output value of $f (x)$ within the interval $7.9 < f (x) < 8.1.$

We also see that we can get output values of $f (x)$ successively closer to 8 by selecting input values closer to 7. In fact, we can obtain output values within any specified interval if we choose appropriate input values.

Figure 4 provides a visual representation of the left- and right-hand limits of the function. From the graph of $f (x),$ we observe the output can get infinitesimally close to $L = 8$ as $x$ approaches 7 from the left and as $x$ approaches 7 from the right.

To indicate the left-hand limit, we write

\lim_{x \to 7^{-}} f (x) = 8.

To indicate the right-hand limit, we write

\lim_{x \to 7^{+}} f (x) = 8.

Graph of the previous function explaining the function's limit at (7, 8) — Figure 4 The left- and right-hand limits are the same for this function.

Understanding Two-Sided Limits

In the previous example, the left-hand limit and right-hand limit as $x$ approaches $a$ are equal. If the left- and right-hand limits are equal, we say that the function $f (x)$ has a two-sided limit as $x$ approaches $a .$ More commonly, we simply refer to a two-sided limit as a limit. If the left-hand limit does not equal the right-hand limit, or if one of them does not exist, we say the limit does not exist.

Finding a Limit Using a Graph

To visually determine if a limit exists as $x$ approaches $a,$ we observe the graph of the function when $x$ is very near to $x = a .$ In Figure 5 we observe the behavior of the graph on both sides of $a .$

Graph of a function that explains the behavior of a limit at (a, L) where the function is increasing when x is less than a and decreasing when x is greater than a. — Figure 5

To determine if a left-hand limit exists, we observe the branch of the graph to the left of $x = a,$ but near $x = a .$ This is where $x < a .$ We see that the outputs are getting close to some real number $L$ so there is a left-hand limit.

To determine if a right-hand limit exists, observe the branch of the graph to the right of $x = a,$ but near $x = a .$ This is where $x > a .$ We see that the outputs are getting close to some real number $L,$ so there is a right-hand limit.

If the left-hand limit and the right-hand limit are the same, as they are in Figure 5, then we know that the function has a two-sided limit. Normally, when we refer to a “limit,” we mean a two-sided limit, unless we call it a one-sided limit.

Finally, we can look for an output value for the function $f (x)$ when the input value $x$ is equal to $a .$ The coordinate pair of the point would be $(a, f (a)) .$ If such a point exists, then $f (a)$ has a value. If the point does not exist, as in Figure 5, then we say that $f (a)$ does not exist.

Example 2

Finding a Limit Using a Graph

Determine the following limits and function value for the function f shown in Figure 6.
1. $\lim_{x \to 2^{-}} f (x)$
2. $\lim_{x \to 2^{+}} f (x)$
3. $\lim_{x \to 2} f (x)$
4. $f (2)$
Figure 6
Determine the following limits and function value for the function f shown in Figure 7.
1. $\lim_{x \to 2^{-}} f (x)$
2. $\lim_{x \to 2^{+}} f (x)$
3. $\lim_{x \to 2} f (x)$
4. $f (2)$
Figure 7

Solution

Looking at Figure 6:
1. $\lim_{x \to 2^{-}} f (x) = 8;$ when $x < 2,$ but infinitesimally close to 2, the output values get close to $y = 8.$
2. $\lim_{x \to 2^{+}} f (x) = 3;$ when $x > 2,$ but infinitesimally close to 2, the output values approach $y = 3.$
3. $\lim_{x \to 2} f (x)$ does not exist because $\lim_{x \to 2^{-}} f (x) \neq \lim_{x \to 2^{+}} f (x);$ the left and right-hand limits are not equal.
4. $f (2) = 3$ because the graph of the function $f$ passes through the point $(2, f (2))$ or $(2, 3) .$
Looking at Figure 7:
1. $\lim_{x \to 2^{-}} f (x) = 8;$ when $x < 2$ but infinitesimally close to 2, the output values approach $y = 8.$
2. $\lim_{x \to 2^{+}} f (x) = 8;$ when $x > 2$ but infinitesimally close to 2, the output values approach $y = 8.$
3. $\lim_{x \to 2} f (x) = 8$ because $\lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{+}} f (x) = 8;$ the left and right-hand limits are equal.
4. $f (2) = 4$ because the graph of the function $f$ passes through the point $(2, f (2))$ or $(2, 4) .$

Finding a Limit Using a Table

Creating a table is a way to determine limits using numeric information. We create a table of values in which the input values of $x$ approach $a$ from both sides. Then we determine if the output values get closer and closer to some real value, the limit $L .$

Let’s consider an example using the following function:

\lim_{x \to 5} (\frac{x^{3} - 125}{x - 5})

To create the table, we evaluate the function at values close to $x = 5.$ We use some input values less than 5 and some values greater than 5 as in Figure 9. The table values show that when $x > 5$ but nearing 5, the corresponding output gets close to 75. When $x > 5$ but nearing 5, the corresponding output also gets close to 75.

Table shows that as x values approach 5 from the positive or negative direction, f(x) gets very close to 75. But when x is equal to 5, y is undefined. — Figure 9

Because

\lim_{x \to 5^{-}} f (x) = 75 = \lim_{x \to 5^{+}} f (x),

then

\lim_{x \to 5} f (x) = 75.

Remember that $f (5)$ does not exist.

Example 3

Finding a Limit Using a Table

Numerically estimate the limit of the following expression by setting up a table of values on both sides of the limit.

\lim_{x \to 0} (\frac{5 \sin (x)}{3 x})

Solution

We can estimate the value of a limit, if it exists, by evaluating the function at values near $x = 0.$ We cannot find a function value for $x = 0$ directly because the result would have a denominator equal to 0, and thus would be undefined.

f (x) = \frac{5 \sin (x)}{3 x}

We create Figure 10 by choosing several input values close to $x = 0,$ with half of them less than $x = 0$ and half of them greater than $x = 0.$ Note that we need to be sure we are using radian mode. We evaluate the function at each input value to complete the table.

The table values indicate that when $x < 0$ but approaching 0, the corresponding output nears $\frac{5}{3} .$

When $x > 0$ but approaching 0, the corresponding output also nears $\frac{5}{3} .$

Table shows that as x values approach 0 from the positive or negative direction, f(x) gets very close to 5 over 3. But when x is equal to 0, y is undefined. — Figure 10

Because

\lim_{x \to 0^{-}} f (x) = \frac{5}{3} = \lim_{x \to 0^{+}} f (x),

then

\lim_{x \to 0} f (x) = \frac{5}{3} .

Example 4

Using a Graphing Utility to Determine a Limit

With the use of a graphing utility, if possible, determine the left- and right-hand limits of the following function as $x$ approaches 0. If the function has a limit as $x$ approaches 0, state it. If not, discuss why there is no limit.

f (x) = 3 \sin (\frac{π}{x})

Solution

We can use a graphing utility to investigate the behavior of the graph close to $x = 0.$ Centering around $x = 0,$ we choose two viewing windows such that the second one is zoomed in closer to $x = 0$ than the first one. The result would resemble Figure 12 for $[- 2, 2]$ by $[- 3, 3] .$

Graph of a sinusodial function zoomed in at [-2, 2] by [-3, 3]. — Figure 12

The result would resemble Figure 13 for $[−0.1, 0.1]$ by $[−3, 3] .$

Graph of the same sinusodial function as in the previous image zoomed in at [-0.1, 0.1] by [-3. 3]. — Figure 13 Even closer to zero, we are even less able to distinguish any limits.

The closer we get to 0, the greater the swings in the output values are. That is not the behavior of a function with either a left-hand limit or a right-hand limit. And if there is no left-hand limit or right-hand limit, there certainly is no limit to the function $f (x)$ as $x$ approaches 0.

We write

\lim_{x \to 0^{-}} (3 \sin (\frac{π}{x})) does not exist .

\lim_{x \to 0^{+}} (3 \sin (\frac{π}{x})) does not exist .

\lim_{x \to 0} (3 \sin (\frac{π}{x})) does not exist .

Key Concepts

A function has a limit if the output values approach some value $L$ as the input values approach some quantity $a .$ See Example 1.
A shorthand notation is used to describe the limit of a function according to the form $\lim_{x \to a} f (x) = L,$ which indicates that as $x$ approaches $a,$ both from the left of $x = a$ and the right of $x = a,$ the output value gets close to $L .$
A function has a left-hand limit if $f (x)$ approaches $L$ as $x$ approaches $a$ where $x < a .$ A function has a right-hand limit if $f (x)$ approaches $L$ as $x$ approaches $a$ where $x > a .$
A two-sided limit exists if the left-hand limit and the right-hand limit of a function are the same. A function is said to have a limit if it has a two-sided limit.
A graph provides a visual method of determining the limit of a function.
If the function has a limit as $x$ approaches $a,$ the branches of the graph will approach the same $y -$ coordinate near $x = a$ from the left and the right. See Example 2.
A table can be used to determine if a function has a limit. The table should show input values that approach $a$ from both directions so that the resulting output values can be evaluated. If the output values approach some number, the function has a limit. See Example 3.
A graphing utility can also be used to find a limit. See Example 4.

Section Exercises

Verbal

Exercise 1

Explain the difference between a value at $x = a$ and the limit as $x$ approaches $a .$

Solution

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

Exercise 2

Explain why we say a function does not have a limit as $x$ approaches $a$ if, as $x$ approaches $a,$ the left-hand limit is not equal to the right-hand limit.

Graphical

For the following exercises, estimate the functional values and the limits from the graph of the function $f$ provided in Figure 14.

A piecewise function with discontinuities at x = -2, x = 1, and x = 4. — Figure 14

Exercise 3

$\lim_{x \to - 2^{-}} f (x)$

Solution

–4

Exercise 4

$\lim_{x \to - 2^{+}} f (x)$

Exercise 5

$\lim_{x \to - 2} f (x)$

Solution

–4

Exercise 6

$f (−2)$

Exercise 7

$\lim_{x \to 1^{-}} f (x)$

Solution

Exercise 8

$\lim_{x \to 1^{+}} f (x)$

Exercise 9

$\lim_{x \to 1} f (x)$

Solution

does not exist

Exercise 10

$f (1)$

Exercise 11

$\lim_{x \to 4^{-}} f (x)$

Solution

Exercise 12

$\lim_{x \to 4^{+}} f (x)$

Exercise 13

$\lim_{x \to 4} f (x)$

Solution

does not exist

Exercise 14

$f (4)$

For the following exercises, draw the graph of a function from the functional values and limits provided.

Exercise 15

$\lim_{x \to 0^{-}} f (x) = 2$ , $\lim_{x \to 0^{+}} f (x) = - 3$ , $\lim_{x \to 2} f (x) = 2$ , $f (0) = 4$ , $f (2) = - 1$ , $f (- 3) does not exist .$

Solution

Answers will vary.

Exercise 16

$\lim_{x \to 2^{-}} f (x) = 0$ , $\lim_{x \to 2^{+}} = - 2$ , $\lim_{x \to 0} f (x) = 3$ , $f (2) = 5$ , $f (0)$

Solution

Answers will vary.

Exercise 17

$\lim_{x \to 2^{-}} f (x) = 2$ , $\lim_{x \to 2^{+}} f (x) = - 3$ , $\lim_{x \to 0} f (x) = 5$ , $f (0) = 1$ , $f (1) = 0$

Solution

Answers will vary.

Exercise 18

$\lim_{x \to 3^{-}} f (x) = 0$ , $\lim_{x \to 3^{+}} f (x) = 5$ , $\lim_{x \to 5} f (x) = 0$ , $f (5) = 4$ , $f (3) does not exist .$

Solution

Answers will vary.

Exercise 19

$\lim_{x \to 4} f (x) = 6$ , $\lim_{x \to 6^{+}} f (x) = - 1$ , $\lim_{x \to 0} f (x) = 5$ , $f (4) = 6$ , $f (2) = 6$

Solution

Answers will vary.

Exercise 20

$\lim_{x \to - 3} f (x) = 2$ , $\lim_{x \to 1^{+}} f (x) = - 2$ , $\lim_{x \to 3} f (x) = - 4$ , $f (- 3) = 0$ , $f (0) = 0$

Solution

Answers will vary.

Exercise 21

$\lim_{x \to π} f (x) = π^{2}$ , $\lim_{x \to - π} f (x) = \frac{π}{2}$ , $\lim_{x \to 1^{-}} f (x) = 0$ , $f (π) = \sqrt{2}$ , $f (0) does not exist .$

Solution

Answers will vary.

For the following exercises, use a graphing calculator to determine the limit to 5 decimal places as $x$ approaches 0.

Exercise 22

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

Exercise 23

$g (x) = {(1 + x)}^{\frac{2}{x}}$

Solution

7.38906

Exercise 24

$h (x) = {(1 + x)}^{\frac{3}{x}}$

Exercise 25

$i (x) = {(1 + x)}^{\frac{4}{x}}$

Solution

54.59815

Exercise 26

$j (x) = {(1 + x)}^{\frac{5}{x}}$

Exercise 27

Based on the pattern you observed in the exercises above, make a conjecture as to the limit of $f (x) = {(1 + x)}^{\frac{6}{x}},$ $g (x) = {(1 + x)}^{\frac{7}{x}},$ $and h (x) = {(1 + x)}^{\frac{n}{x}} .$

Solution

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

For the following exercises, use a graphing utility to find graphical evidence to determine the left- and right-hand limits of the function given as $x$ approaches $a .$ If the function has a limit as $x$ approaches $a,$ state it. If not, discuss why there is no limit.

Exercise 28

$(x) = {\begin{array}{l} | x | - 1, & if x \neq 1 \\ x^{3}, & if x = 1 \end{array} a = 1$

Exercise 29

$(x) = {\begin{array}{l} \frac{1}{x + 1}, & if x = - 2 \\ {(x + 1)}^{2}, & if x \neq - 2 \end{array} a = - 2$

Solution

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

Numeric

For the following exercises, use numerical evidence to determine whether the limit exists at $x = a .$ If not, describe the behavior of the graph of the function near $x = a .$ Round answers to two decimal places.

Exercise 30

$f (x) = \frac{x^{2} - 4 x}{16 - x^{2}}; a = 4$

Exercise 31

$f (x) = \frac{x^{2} - x - 6}{x^{2} - 9}; a = 3$

Solution

$\lim_{x \to 3} (\frac{x^{2} - x - 6}{x^{2} - 9}) = \frac{5}{6} \approx 0.83$

Exercise 32

$f (x) = \frac{x^{2} - 6 x - 7}{x^{2} - 7 x}; a = 7$

Exercise 33

$f (x) = \frac{x^{2} - 1}{x^{2} - 3 x + 2}; a = 1$

Solution

$\lim_{x \to 1} (\frac{x^{2} - 1}{x^{2} - 3 x + 2}) = - 2.00$

Exercise 34

$f (x) = \frac{1 - x^{2}}{x^{2} - 3 x + 2}; a = 1$

Exercise 35

$f (x) = \frac{10 - 10 x^{2}}{x^{2} - 3 x + 2}; a = 1$

Solution

$\lim_{x \to 1} (\frac{10 - 10 x^{2}}{x^{2} - 3 x + 2}) = 20.00$

Exercise 36

$f (x) = \frac{x}{6 x^{2} - 5 x - 6}; a = \frac{3}{2}$

Exercise 37

$f (x) = \frac{x}{4 x^{2} + 4 x + 1}; a = - \frac{1}{2}$

Solution

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

Exercise 38

$f (x) = \frac{2}{x - 4}; a = 4$

For the following exercises, use a calculator to estimate the limit by preparing a table of values. If there is no limit, describe the behavior of the function as $x$ approaches the given value.

Exercise 39

$\lim_{x \to 0} \frac{7 \tan x}{3 x}$

Solution

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

Table shows as the function approaches 0, the value is 7 over 3 but the function is undefined at 0.

Exercise 40

$\lim_{x \to 4} \frac{x^{2}}{x - 4}$

Solution

Table shows as the function approaches 4, the value does not exist since approaching the limit value from the left is negative infinity and approaching the limit value from the right is positive infinity.

Exercise 41

$\lim_{x \to 0} \frac{2 \sin x}{4 \tan x}$

For the following exercises, use a graphing utility to find numerical or graphical evidence to determine the left and right-hand limits of the function given as $x$ approaches $a .$ If the function has a limit as $x$ approaches $a,$ state it. If not, discuss why there is no limit.

Exercise 42

$\lim_{x \to 0} e^{e^{\frac{1}{x}}}$

Exercise 43

$\lim_{x \to 0} e^{e^{- \frac{1}{x^{2}}}}$

Solution

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

Exercise 44

$\lim_{x \to 0} \frac{| x |}{x}$

Exercise 45

$\lim_{x \to - 1} \frac{| x + 1 |}{x + 1}$

Solution

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

Exercise 46

$\lim_{x \to 5} \frac{| x - 5 |}{5 - x}$

Exercise 47

$\lim_{x \to - 1} \frac{1}{{(x + 1)}^{2}}$

Solution

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

Exercise 48

$\lim_{x \to 1} \frac{1}{{(x - 1)}^{3}}$

Exercise 49

$\lim_{x \to 0} \frac{5}{1 - e^{\frac{2}{x}}}$

Solution

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

Exercise 50

Use numerical and graphical evidence to compare and contrast the limits of two functions whose formulas appear similar: $f (x) = | \frac{1 - x}{x} |$ and $g (x) = | \frac{1 + x}{x} |$ as $x$ approaches 0. Use a graphing utility, if possible, to determine the left- and right-hand limits of the functions $f (x)$ and $g (x)$ as $x$ approaches 0. If the functions have a limit as $x$ approaches 0, state it. If not, discuss why there is no limit.

Extensions

Exercise 51

According to the Theory of Relativity, the mass $m$ of a particle depends on its velocity $v$ . That is

m = \frac{m_{o}}{\sqrt{1 - (v^{2} / c^{2})}}

where $m_{o}$ is the mass when the particle is at rest and $c$ is the speed of light. Find the limit of the mass, $m,$ as $v$ approaches $c^{-} .$

Solution

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,

\lim_{v \to c^{-}} m = \lim_{v \to c^{-}} \frac{m_{o}}{\sqrt{1 - (v^{2} / c^{2})}} = \infty

Exercise 52

Allow the speed of light, $c,$ to be equal to 1.0. If the mass, $m,$ is 1, what occurs to $m$ as $v \to c ?$ Using the values listed in Table 1, make a conjecture as to what the mass is as $v$ approaches 1.00.

Table 1
$v$	$m$
0.5	1.15
0.9	2.29
0.95	3.20
0.99	7.09
0.999	22.36
0.99999	223.61

left-hand limit

the limit of values of

f (x)

x

approaches from

a

the left, denoted

\lim_{x \to a^{-}} f (x) = L .

The values of

f (x)

can get as close to the limit

L

as we like by taking values of

x

sufficiently close to

a

such that

x < a

and

x \neq a .

Both

a

and

L

are real numbers.

limit

when it exists, the value,

L,

that the output of a function

f (x)

approaches as the input

x

gets closer and closer to

a

but does not equal

a .

The value of the output,

f (x),

can get as close to

L

as we choose to make it by using input values of

x

sufficiently near to

x = a,

but not necessarily at

x = a .

Both

a

and

L

are real numbers, and

L

is denoted

\lim_{x \to a} f (x) = L .

right-hand limit

the limit of values of

f (x)

x

approaches

a

from the right, denoted

\lim_{x \to a^{+}} f (x) = L .

The values of

f (x)

can get as close to the limit

L

as we like by taking values of

x

sufficiently close to

a

where

x > a,

and

x \neq a .

Both

a

and

L

are real numbers.

two-sided limit

the limit of a function

f (x),

x

approaches

a,

is equal to

L,

that is,

\lim_{x \to a} f (x) = L

if and only if

\lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) .