Finding Limits: Properties of Limits

Consider the rational function

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

The function can be factored as follows:

f (x) = \frac{(x - 7) (x + 1)}{x - 7}, which gives us f (x) = x + 1, x \neq 7.

Does this mean the function $f$ is the same as the function $g (x) = x + 1 ?$

The answer is no. Function $f$ does not have $x = 7$ in its domain, but $g$ does. Graphically, we observe there is a hole in the graph of $f (x)$ at $x = 7,$ as shown in Figure 1 and no such hole in the graph of $g (x),$ as shown in Figure 2.

Graph of an increasing function where f(x) = (x^2-6x-7)\(x-7) with a discontinuity at (7, 8) — Figure 1 The graph of function $f$ contains a break at $x = 7$ and is therefore not continuous at $x = 7.$

Graph of an increasing function where g(x) = x+1 — Figure 2 The graph of function $g$ is continuous.

So, do these two different functions also have different limits as $x$ approaches 7?

Not necessarily. Remember, in determining a limit of a function as $x$ approaches $a,$ what matters is whether the output approaches a real number as we get close to $x = a .$ The existence of a limit does not depend on what happens when $x$ equals $a .$

Look again at Figure 1 and Figure 2. Notice that in both graphs, as $x$ approaches 7, the output values approach 8. This means

\lim_{x \to 7} f (x) = \lim_{x \to 7} g (x) .

Remember that when determining a limit, the concern is what occurs near $x = a,$ not at $x = a .$ In this section, we will use a variety of methods, such as rewriting functions by factoring, to evaluate the limit. These methods will give us formal verification for what we formerly accomplished by intuition.

Finding the Limit of a Sum, a Difference, and a Product

Graphing a function or exploring a table of values to determine a limit can be cumbersome and time-consuming. When possible, it is more efficient to use the properties of limits, which is a collection of theorems for finding limits.

Knowing the properties of limits allows us to compute limits directly. We can add, subtract, multiply, and divide the limits of functions as if we were performing the operations on the functions themselves to find the limit of the result. Similarly, we can find the limit of a function raised to a power by raising the limit to that power. We can also find the limit of the root of a function by taking the root of the limit. Using these operations on limits, we can find the limits of more complex functions by finding the limits of their simpler component functions.

Properties of Limits

Let $a, k, A,$ and $B$ represent real numbers, and $f$ and $g$ be functions, such that $\lim_{x \to a} f (x) = A$ and $\lim_{x \to a} g (x) = B .$ For limits that exist and are finite, the properties of limits are summarized in Table 1

Table 1 ..
Constant, k	$\lim_{x \to a} k = k$
Constant times a function	$\lim_{x \to a} [k \cdot f (x)] = k \lim_{x \to a} f (x) = k A$
Sum of functions	$\lim_{x \to a} [f (x) + g (x)] = \lim_{x \to a} f (x) + \lim_{x \to a} g (x) = A + B$
Difference of functions	$\lim_{x \to a} [f (x) - g (x)] = \lim_{x \to a} f (x) - \lim_{x \to a} g (x) = A - B$
Product of functions	$\lim_{x \to a} [f (x) \cdot g (x)] = \lim_{x \to a} f (x) \cdot \lim_{x \to a} g (x) = A \cdot B$
Quotient of functions	$\lim_{x \to a} \frac{f (x)}{g (x)} = \frac{\lim_{x \to a} f (x)}{\lim_{x \to a} g (x)} = \frac{A}{B}, B \neq 0$
Function raised to an exponent	$\lim_{x \to a} {[f (x)]}^{n} = {[\lim_{x \to a} f (x)]}^{n} = A^{n},$ where $n$ is a positive integer
nth root of a function, where n is a positive integer	$\lim_{x \to a} \sqrt[n]{f (x)} = \sqrt[n]{\lim_{x \to a} [f (x)]} = \sqrt[n]{A}$
Polynomial function	$\lim_{x \to a} p (x) = p (a)$

Example 1

Evaluating the Limit of a Function Algebraically

Evaluate $\lim_{x \to 3} (2 x + 5) .$

Solution

\begin{array}{l} \lim_{x \to 3} (2 x + 5) = \lim_{x \to 3} (2 x) + \lim_{x \to 3} (5) & Sum of functions property \\ = \underset{x \to 3}{2 \lim} (x) + \lim_{x \to 3} (5) & Constant times a function property \\ = 2 (3) + 5 & Evaluate \\ = 11 \end{array}

Finding the Limit of a Polynomial

Not all functions or their limits involve simple addition, subtraction, or multiplication. Some may include polynomials. Recall that a polynomial is an expression consisting of the sum of two or more terms, each of which consists of a constant and a variable raised to a nonnegative integral power. To find the limit of a polynomial function, we can find the limits of the individual terms of the function, and then add them together. Also, the limit of a polynomial function as $x$ approaches $a$ is equivalent to simply evaluating the function for $a$ .

Example 2

Evaluating the Limit of a Function Algebraically

Evaluate $\lim_{x \to 3} (5 x^{2}) .$

Solution

\begin{array}{l} \lim_{x \to 3} (5 x^{2}) = 5 \lim_{x \to 3} (x^{2}) & Constant times a function property \\ = 5 (3^{2}) & Function raised to an exponent property \\ = 45 \end{array}

Example 3

Evaluating the Limit of a Polynomial Algebraically

Evaluate $\lim_{x \to 5} (2 x^{3} - 3 x + 1) .$

Solution

\begin{array}{l} \lim_{x \to 5} (2 x^{3} - 3 x + 1) = \lim_{x \to 5} (2 x^{3}) - \lim_{x \to 5} (3 x) + \lim_{x \to 5} (1) & Sum of functions \\ = \underset{x \to 5}{2 \lim} (x^{3}) - \underset{x \to 5}{3 \lim} (x) + \lim_{x \to 5} (1) & Constant times a function \\ = 2 (5^{3}) - 3 (5) + 1 & Function raised to an exponent \\ = 236 & Evaluate \end{array}

Finding the Limit of a Power or a Root

When a limit includes a power or a root, we need another property to help us evaluate it. The square of the limit of a function equals the limit of the square of the function; the same goes for higher powers. Likewise, the square root of the limit of a function equals the limit of the square root of the function; the same holds true for higher roots.

Example 4

Evaluating a Limit of a Power

Evaluate $\lim_{x \to 2} {(3 x + 1)}^{5} .$

Solution

We will take the limit of the function as $x$ approaches 2 and raise the result to the 5^th power.

\begin{array}{l} \lim_{x \to 2} {(3 x + 1)}^{5} = {(\lim_{x \to 2} (3 x + 1))}^{5} \\ = {(3 (2) + 1)}^{5} \\ = 7^{5} \\ = 16,807 \end{array}

Finding the Limit of a Quotient

Finding the limit of a function expressed as a quotient can be more complicated. We often need to rewrite the function algebraically before applying the properties of a limit. If the denominator evaluates to 0 when we apply the properties of a limit directly, we must rewrite the quotient in a different form. One approach is to write the quotient in factored form and simplify.

Example 5

Evaluating the Limit of a Quotient by Factoring

Evaluate $\lim_{x \to 2} (\frac{x^{2} - 6 x + 8}{x - 2}) .$

Solution

Factor where possible, and simplify.

\begin{array}{l} \lim_{x \to 2} (\frac{x^{2} - 6 x + 8}{x - 2}) = \lim_{x \to 2} (\frac{(x - 2) (x - 4)}{x - 2}) & Factor the numerator . \\ = \lim_{x \to 2} (\frac{(x - 2) (x - 4)}{x - 2}) & Cancel the common factors . \\ = \lim_{x \to 2} (x - 4) & Evaluate . \\ = 2 - 4 = - 2 \end{array}

Analysis

When the limit of a rational function cannot be evaluated directly, factored forms of the numerator and denominator may simplify to a result that can be evaluated.

Notice, the function

f (x) = \frac{x^{2} - 6 x + 8}{x - 2}

is equivalent to the function

f (x) = x - 4, x \neq 2.

Notice that the limit exists even though the function is not defined at $x = 2 .$

Example 6

Evaluating the Limit of a Quotient by Finding the LCD

Evaluate $\lim_{x \to 5} (\frac{\frac{1}{x} - \frac{1}{5}}{x - 5}) .$

Solution

Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have the LCD as their denominator.

Analysis

When determining the limit of a rational function that has terms added or subtracted in either the numerator or denominator, the first step is to find the common denominator of the added or subtracted terms; then, convert both terms to have that denominator, or simplify the rational function by multiplying numerator and denominator by the least common denominator. Then check to see if the resulting numerator and denominator have any common factors.

Example 7

Evaluating a Limit Containing a Root Using a Conjugate

Evaluate $\lim_{x \to 0} (\frac{\sqrt{25 - x} - 5}{x}) .$

Solution

$\begin{array}{l} \lim_{x \to 0} (\frac{\sqrt{25 - x} - 5}{x}) = \lim_{x \to 0} (\frac{(\sqrt{25 - x} - 5)}{x} \cdot \frac{(\sqrt{25 - x} + 5)}{(\sqrt{25 - x} + 5)}) & Multiply numerator and denominator by the conjugate . \\ = \lim_{x \to 0} (\frac{(25 - x) - 25}{x (\sqrt{25 - x} + 5)}) & Multiply: (\sqrt{25 - x} - 5) \cdot (\sqrt{25 - x} + 5) = (25 - x) - 25. \\ = \lim_{x \to 0} (\frac{- x}{x (\sqrt{25 - x} + 5)}) & Combine like terms . \\ = \lim_{x \to 0} (\frac{- x}{x (\sqrt{25 - x} + 5)}) & Simplify \frac{- x}{x} = - 1. \\ = \frac{- 1}{\sqrt{25 - 0} + 5} & Evaluate . \\ = \frac{- 1}{5 + 5} = - \frac{1}{10} \end{array}$

Analysis

When determining a limit of a function with a root as one of two terms where we cannot evaluate directly, think about multiplying the numerator and denominator by the conjugate of the terms.

Example 8

Evaluating the Limit of a Quotient of a Function by Factoring

Evaluate $\lim_{x \to 4} (\frac{4 - x}{\sqrt{x} - 2}) .$

Solution

$\begin{array}{l} \lim_{x \to 4} (\frac{4 - x}{\sqrt{x} - 2}) = \lim_{x \to 4} (\frac{(2 + \sqrt{x}) (2 - \sqrt{x})}{\sqrt{x} - 2}) & Factor. \\ = \lim_{x \to 4} (\frac{(2 + \sqrt{x}) (2 - \sqrt{x})}{- (2 - \sqrt{x})}) & Factor −1 out of the denominator . Simplify . \\ = \lim_{x \to 4} - (2 + \sqrt{x}) & Evaluate . \\ = - (2 + \sqrt{4}) \\ = - 4 \end{array}$

Analysis

Multiplying by a conjugate would expand the numerator; look instead for factors in the numerator. Four is a perfect square so that the numerator is in the form

a^{2} - b^{2}

and may be factored as

(a + b) (a - b) .

Example 9

Evaluating the Limit of a Quotient with Absolute Values

Evaluate $\lim_{x \to 7} \frac{| x - 7 |}{x - 7} .$

Solution

The function is undefined at $x = 7,$ so we will try values close to 7 from the left and the right.

Left-hand limit: $\frac{| 6.9 - 7 |}{6.9 - 7} = \frac{| 6.99 - 7 |}{6.99 - 7} = \frac{| 6.999 - 7 |}{6.999 - 7} = - 1$

Right-hand limit: $\frac{| 7.1 - 7 |}{7.1 - 7} = \frac{| 7.01 - 7 |}{7.01 - 7} = \frac{| 7.001 - 7 |}{7.001 - 7} = 1$

Since the left- and right-hand limits are not equal, there is no limit.

Key Concepts

The properties of limits can be used to perform operations on the limits of functions rather than the functions themselves. See Example 1.
The limit of a polynomial function can be found by finding the sum of the limits of the individual terms. See Example 2 and Example 3.
The limit of a function that has been raised to a power equals the same power of the limit of the function. Another method is direct substitution. See Example 4.
The limit of the root of a function equals the corresponding root of the limit of the function.
One way to find the limit of a function expressed as a quotient is to write the quotient in factored form and simplify. See Example 5.
Another method of finding the limit of a complex fraction is to find the LCD. See Example 6.
A limit containing a function containing a root may be evaluated using a conjugate. See Example 7.
The limits of some functions expressed as quotients can be found by factoring. See Example 8.
One way to evaluate the limit of a quotient containing absolute values is by using numeric evidence. Setting it up piecewise can also be useful. See Example 9.

Section Exercises

Verbal

Exercise 1

Give an example of a type of function $f$ whose limit, as $x$ approaches $a,$ is $f (a) .$

Solution

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

Exercise 2

When direct substitution is used to evaluate the limit of a rational function as $x$ approaches $a$ and the result is $f (a) = \frac{0}{0},$ does this mean that the limit of $f$ does not exist?

Exercise 3

What does it mean to say the limit of $f (x),$ as $x$ approaches $c,$ is undefined?

Solution

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.

Algebraic

For the following exercises, evaluate the limits algebraically.

Exercise 4

$\lim_{x \to 0} (3)$

Exercise 5

$\lim_{x \to 2} (\frac{- 5 x}{x^{2} - 1})$

Solution

$\frac{- 10}{3}$

Exercise 6

$\lim_{x \to 2} (\frac{x^{2} - 5 x + 6}{x + 2})$

Exercise 7

$\lim_{x \to 3} (\frac{x^{2} - 9}{x - 3})$

Solution

Exercise 8

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

Exercise 9

$\lim_{x \to \frac{3}{2}} (\frac{6 x^{2} - 17 x + 12}{2 x - 3})$

Solution

$\frac{1}{2}$

Exercise 10

$\lim_{x \to - \frac{7}{2}} (\frac{8 x^{2} + 18 x - 35}{2 x + 7})$

Exercise 11

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

Solution

Exercise 12

$\lim_{x \to - 3} (\frac{- 7 x^{4} - 21 x^{3}}{- 12 x^{4} + 108 x^{2}})$

Exercise 13

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

Solution

does not exist

Exercise 14

$\lim_{h \to 0} (\frac{{(3 + h)}^{3} - 27}{h})$

Exercise 15

$\lim_{h \to 0} (\frac{{(2 - h)}^{3} - 8}{h})$

Solution

$- 12$

Exercise 16

$\lim_{h \to 0} (\frac{{(h + 3)}^{2} - 9}{h})$

Exercise 17

$\lim_{h \to 0} (\frac{\sqrt{5 - h} - \sqrt{5}}{h})$

Solution

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

Exercise 18

$\lim_{x \to 0} (\frac{\sqrt{3 - x} - \sqrt{3}}{x})$

Exercise 19

$\lim_{x \to 9} (\frac{x^{2} - 81}{3 - \sqrt{x}})$

Solution

$- 108$

Exercise 20

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

Exercise 21

$\lim_{x \to 0} (\frac{x}{\sqrt{1 + 2 x} - 1})$

Solution

Exercise 22

$\lim_{x \to \frac{1}{2}} (\frac{x^{2} - \frac{1}{4}}{2 x - 1})$

Exercise 23

$\lim_{x \to 4} (\frac{x^{3} - 64}{x^{2} - 16})$

Solution

Exercise 24

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

Exercise 25

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

Solution

Exercise 26

$\lim_{x \to 2} (\frac{| x - 2 |}{x - 2})$

Exercise 27

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

Solution

Exercise 28

$\lim_{x \to 4^{+}} (\frac{| x - 4 |}{4 - x})$

Exercise 29

$\lim_{x \to 4} (\frac{| x - 4 |}{4 - x})$

Solution

does not exist

Exercise 30

$\lim_{x \to 2} (\frac{- 8 + 6 x - x^{2}}{x - 2})$

For the following exercise, use the given information to evaluate the limits: $\lim_{x \to c} f (x) = 3,$ $\lim_{x \to c} g (x) = 5$ .

Exercise 31

$\lim_{x \to c} [2 f (x) + \sqrt{g (x)}]$

Solution

$6 + \sqrt{5}$

Exercise 32

$\lim_{x \to c} [3 f (x) + \sqrt{g (x)}]$

Exercise 33

$\lim_{x \to c} \frac{f (x)}{g (x)}$

Solution

$\frac{3}{5}$

For the following exercises, evaluate the following limits.

Exercise 34

$\lim_{x \to 2} \cos (π x)$

Exercise 35

$\lim_{x \to 2} \sin (π x)$

Solution

Exercise 36

$\lim_{x \to 2} \sin (\frac{π}{x})$

Exercise 37

$f (x) = {\begin{array}{l} 2 x^{2} + 2 x + 1, & x \leq 0 \\ x - 3, & x > 0 \end{array}; \lim_{x \to 0^{+}} f (x)$

Solution

$- 3$

Exercise 38

$f (x) = {\begin{array}{l} 2 x^{2} + 2 x + 1, & x \leq 0 \\ x - 3, & x > 0 \end{array}; \lim_{x \to 0^{-}} f (x)$

Exercise 39

$f (x) = {\begin{array}{l} 2 x^{2} + 2 x + 1, & x \leq 0 \\ x - 3, & x > 0 \end{array}; \lim_{x \to 0} f (x)$

Solution

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

Exercise 40

$\lim_{x \to 4} \frac{\sqrt{x + 5} - 3}{x - 4}$

Exercise 41

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

Solution

Exercise 42

$\lim_{x \to 2} \frac{\sqrt{x + 7} - 3}{x^{2} - x - 2}$

Exercise 43

$\lim_{x \to 3^{+}} \frac{x^{2}}{x^{2} - 9}$

Solution

Limit does not exist; limit approaches infinity.

For the following exercises, find the average rate of change $\frac{f (x + h) - f (x)}{h} .$

Exercise 44

$f (x) = x + 1$

Exercise 45

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

Solution

$4 x + 2 h$

Exercise 46

$f (x) = x^{2} + 3 x + 4$

Exercise 47

$f (x) = x^{2} + 4 x - 100$

Solution

$2 x + h + 4$

Exercise 48

$f (x) = 3 x^{2} + 1$

Exercise 49

$f (x) = \cos (x)$

Solution

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

Exercise 50

$f (x) = 2 x^{3} - 4 x$

Exercise 51

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

Solution

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

Exercise 52

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

Exercise 53

$f (x) = \sqrt{x}$

Solution

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

Graphical

Exercise 54

Find an equation that could be represented by Figure 3.

Graph of increasing function with a removable discontinuity at (2, 3). — Figure 3

Exercise 55

Find an equation that could be represented by Figure 4.

Graph of increasing function with a removable discontinuity at (-3, -1). — Figure 4

Solution

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

For the following exercises, refer to Figure 5.

Graph of increasing function from zero to positive infinity. — Figure 5

Exercise 56

What is the right-hand limit of the function as $x$ approaches 0?

Exercise 57

What is the left-hand limit of the function as $x$ approaches 0?

Solution

does not exist

Real-World Applications

Exercise 58

The position function $s (t) = - 16 t^{2} + 144 t$ gives the position of a projectile as a function of time. Find the average velocity (average rate of change) on the interval $[1, 2]$ .

Exercise 59

The height of a projectile is given by $s (t) = - 64 t^{2} + 192 t$ Find the average rate of change of the height from $t = 1$ second to $t = 1.5$ seconds.

Solution

Exercise 60

The amount of money in an account after $t$ years compounded continuously at 4.25% interest is given by the formula $A = A_{0} e^{0.0425 t},$ where $A_{0}$ is the initial amount invested. Find the average rate of change of the balance of the account from $t = 1$ year to $t = 2$ years if the initial amount invested is $1,000.00.

properties of limits

a collection of theorems for finding limits of functions by performing mathematical operations on the limits