Lesson Explainer: Evaluating Limits Using Algebraic Techniques

In this explainer, we will learn how to use algebraic techniques like factorization to evaluate limits.

The limit of a function at a point describes the behavior of the function near the given point, rather than the value of the function at the point. While this distinction is very important to keep in mind, we know that, in many cases, the limit of a function is actually equal to the value of the function at the point. In such cases, we can evaluate the limit by directly substituting the limit point into the function.

This method is clearly invalid if the function is not defined at the limit point. In other words, if the limit point does not belong to the domain of the function, we cannot evaluate the limit by direct substitution. We can still use a table of values or the graph of the function to estimate the limit, but these methods usually require a calculator to complete and often do not give us the full idea of the value of a limit. In this explainer, we will introduce an algebraic method of evaluating certain limits without using calculators.

In all of the limits that we will consider in this explainer, the functions will be in the form of a quotient so that the denominator is equal to zero at the limit point, and thus the limit point will not lie in the domain. More specifically, we consider limits of the form

We cannot find such limits by direct substitution since substituting the limit point into the quotient will result in having a zero in the denominator. If the numerator does not approach zero near the limit point (i.e., ), then we can say that this limit does not exist. This is because the quotient will become larger and larger in size (absolute value) if the denominator approaches zero while the numerator does not. We can understand this better by considering a special case where and . We can compute a table of function values near the limit point .

					0	0.01	0.1	0.5	1
					undefined	100	10	2	1

From this table, we can clearly see that does not approach a particular value as approaches 0. In fact, the absolute value of gets larger as gets closer to 0. Thus, we can conclude that does not exist. We can also observe this in the graph of .

In the graph above, we can see that the absolute value of the function increases without bound as approaches 0. In general, this is what we expect from the limit of the form where the denominator approaches 0 as approaches but the numerator does not.

Property: Limit of a Quotient

The limit does not exist if both of the following are true:

,
or does not exist (i.e., there is no value that approaches as approaches ).

Hence, when evaluating the limit of a quotient, we should first check whether the denominator approaches zero. If it does, then we should check whether the numerator also approaches zero. If it does not, we know that the limit does not exist, denoted DNE. We can remember this by writing symbolically

Here, the numerator 1 is symbolic of any nonzero constant, and it does not mean that the numerator has to be equal to 1.

This leaves us with the other case where the numerator also approaches zero. Formally, this leads to the case , which is called an indeterminate form.

Definition: Indeterminate Form

An indeterminate form is an algebraic expression of numbers or infinity whose behavior cannot be determined in the current form. For instance, the expression is an indeterminate form.

Indeterminate forms occur often when evaluating limits, but it is very important to keep in mind that an indeterminate form is never the final answer. Instead, indeterminate forms mean that we cannot determine the value of the limit using that specific method. In particular, if and both approach zero as approaches , we can say that the limit takes the indeterminate form . But it is incorrect to state that this limit equals . It cannot be overemphasized that the indeterminate form is not an answer to the limit problem. It also does not mean that the limit does not exist or that we are unable to find the limit. When we notice that our limit takes an indeterminate form, we need to find a different method to find this limit.

In our first example, we will consider a limit that takes the indeterminate form .

Example 1: Finding the Limit of a Rational Function at a Point

Find .

Answer

In this example, we need to find the limit of a rational function. We know that we can find the limit of a rational function by direct substitution only if the denominator does not equal zero at the limit point. We can check this condition by substituting the limit point into the denominator :

Since the denominator is equal to zero at the limit point, we cannot find this limit by direct substitution. In such a case, we know that the limit does not exist if the numerator is not also equal to zero. We can check this condition by substituting into the numerator:

This means that our quotient takes the form when we substitute the limit point. Recall that is an indeterminate form, and it is not a valid answer to any limit problem. This tells us we must use a different method to evaluate this limit.

We noted that both the numerator and denominator are equal to zero at . We recall the remainder theorem, which states that if is a polynomial with , then is a factor of . In this case, the remainder theorem tells us that is a factor of the polynomials in the numerator and denominator of the quotient. This means that after factoring both polynomials, we can cancel this factor from the quotient. Canceling this factor from the two polynomials means that unless this is a repeated factor of one of the polynomials, the resulting polynomials will no longer have 4 as their root. This means that substituting into the simplified quotient will not produce a zero in the denominator, hence validating the direct substitution method in the limit.

Let us work out this simplification by first factoring the numerator. Noting the common factor 3 in the numerator, we can write

To factor the denominator, we recall the difference between squares formula, . Using this formula, we can write

Hence, we can write the quotient as

We see that the factor is common between both the numerator and denominator of this quotient. Canceling out this common factor, we can write this function as

Note that this new function has the same value as the original rational function, except at the point where the original function is undefined. Since the limit of a function only concerns the value of the function near the limit point, we know that this function has the same limit at as the original rational function. In other words,

The limit on the right-hand side of the equation above is the limit of a rational function where the denominator is not equal to zero at the limit point. Hence, we can solve this limit by direct substitution:

Hence, we have

In the previous example, we found the limit of a function in the indeterminate form . Although the original limit was in the indeterminate form, we were able to algebraically simplify the given function and then use direct substitution to finish the problem. The main idea of the algebraic method to solve such a limit is to simplify this quotient via algebraic manipulations with another expression where we will be able to use the direct substitution method.

When we simplify a rational function by canceling a common factor in the numerator and denominator, the resulting function is usually not identical to the initial one. In particular, the domain of the resulting expression is often larger by one number that does not belong to the domain of the original function. The enlarged domain in the simplified function is what makes it possible to use direct substitution to evaluate the limit.

In the graph of the original function, this limit point is represented by an open circle, indicating that the function is undefined at this point. On the other hand, the graph of the simplified function is identical to that of the original function, except that it does not have an empty hole. We can see this difference in the graphs of the original and simplified functions from the previous example.

From the graph, we can see that the original function is not defined at , while the simplified function is defined there. Otherwise, the two graphs are identical. Since the limit of a function does not concern the value of the function at the limit point, this difference does not affect the limit. Hence, the limit of the original function can be obtained by finding the limit of the simplified function, which can be obtained by direct substitution.

Property: Limits of Functions

Let and be functions satisfying for any . Then,

This property can justify most of the algebraic steps to evaluate limits. We will assume this property when we are using algebraic methods to evaluate limits.

How To: Evaluating Limits of Rational Functions in the Indeterminate Form 0/0

Let and be polynomial functions satisfying and . Then, to evaluate the limit of the form , we need to

factor both and ,
cancel out all common factors,
equate the limit of the simplified quotient to the original limit,
evaluate the limit.

If two polynomials and satisfy the conditions and , they must both have factors by the remainder theorem. Then, these factors will cancel out and result in a simplified rational function. If this step eliminates all of the factors from the denominator, the simplified quotient will no longer have a zero denominator at the limit point . In this case, we can evaluate the limit using direct substitution. On the other hand, if this step eliminates all of the factors from the numerator but not from the denominator, the resulting limit would be in the form , which means that the limit would not exist.

Let us consider another example of evaluating the limit of a rational function in the indeterminate form using algebraic methods.

Example 2: Finding the Limit of a Rational Function at a Point by Factoring

Find .

Answer

In this example, we need to find the limit of a rational function. We know that we can find the limit of a rational function by direct substitution only if the denominator does not equal zero at the limit point. Let us begin by examining what happens to the given quotient when we substitute the limit point :

The expression is called an indeterminate form. We know that an indeterminate form is never a valid answer to a limit problem. Recall that we can attempt to find the limit of a rational function in indeterminate form by following the steps below:

Factor both the numerator and denominator.
Cancel out all common factors.
Equate the limit of the simplified quotient to the original limit.
Evaluate the limit.

We can begin by noting the common factor 2 in the denominator, which leads to

We can factor the quadratic expression in the parentheses to write the denominator as

Next, we factor the numerator, which is a cubic polynomial. Noting that , we can write

This is a sum of two cubes, which can be factored using the formula

Applying this formula to the numerator with and , we obtain

This leads to

Thus, we have

In the next example, we can apply the algebraic method to find the limit of a rational function in the indeterminate form under the square root.

Example 3: Finding the Limit of a Composition of Rational and Root Functions at a Point

Find .

Answer

In this example, we need to find the limit of a function that is a composition of the square root function and a rational function. We know that we can find the limit of a function that is a composition of a rational function and the square root function by direct substitution if the limit point belongs to the domain of the function. For a number to belong to the domain of such a function, we must make sure that the denominator of the rational function does not equal zero at this number and that the expression under the square root is nonnegative at this number. In short, if we are able to compute the value of the function at the given number, then the number belongs to the domain.

Let us begin by examining what happens to the given function when we substitute the limit point :

Factor both the numerator and denominator.
Cancel out all common factors.
Equate the limit of the simplified quotient to the original limit.
Evaluate the limit.

Although the function in our limit is not strictly a rational function, the same simplification step in the rational function under the square root may be used to find the limit.

Factoring the numerator of the quotient,

The denominator is factored as

Then, we can simplify

Canceling out the factor in the last step is significant because this factor is responsible for making both the numerator and denominator equal to zero at the limit point. We can see that the denominator of the simplified rational function under the square root is no longer equal to zero at the limit point . Also, the quotient under the square root takes a positive value at the limit point. This means that is in the domain of this function that is a composition of the square root function and a rational function. We know that, in such cases, we can compute the limit by direct substitution. This leads to

Hence,

In previous examples, we found the limit of a rational function in indeterminate form by factoring the numerator and denominator of the function and simplifying. We noted that when the limit of a rational function at results in an indeterminate form , there is always a factor in both the numerator and denominator. This means that we can always simplify this rational function until either the numerator or the denominator (in most cases both) is nonzero at the limit point. At that point, we can find the limit by direct substitution.

However, this process can be more involved if we encounter a function that is difficult to factor. The saving grace in such situations is knowing that there has to be a factor of the form in both the numerator and denominator. Knowing this means that we can find a factored form of the polynomial as for some other polynomial . This process is done using the polynomial long division, as demonstrated in the next example.

Example 4: Finding the Limit of a Rational Function at a Point

Find .

Answer

In this example, we need to find the limit of a rational function. We know that we can find the limit of a rational function by direct substitution only if the denominator does not equal zero at the limit point. Let us begin by examining what happens to the given quotient when we substitute the limit point :

Factor both the numerator and denominator.
Cancel out all common factors.
Equate the limit of the simplified quotient to the original limit.
Evaluate the limit.

In particular, we know that both the numerator and denominator contain the factor since these polynomials are equal to zero at . Let us begin by factoring the numerator. We know that , so we can use the difference of squares formula, , to factor the numerator:

While the numerator is simple to factor, the denominator is more involved. To find a factorization of the denominator in the form , we will use the polynomial long division. Dividing the denominator by ,

This leads to the factorization of the denominator

This leads to

Canceling out the factor in the last step is significant because this factor is responsible for making both the numerator and denominator equal to zero at the limit point. If the denominator of the simplified quotient does not equal zero at the limit point , we can find this limit by direct substitution:

Hence,

So far, we learned how to evaluate the limit of a rational function in the indeterminate form using algebraic methods. We now turn our focus to limits in an indeterminate form where the numerator or the denominator has square roots. For rational functions, we could solve the problem of the indeterminate form by factoring the numerator and denominator and canceling out the common factor. This is not possible in general for quotients that are not rational functions, since it is difficult to factor expressions that are not polynomials.

In the case where the quotient contains square root expressions, a trick using a conjugate expression is useful. A pair of expressions and are called a conjugate pair. Using the difference of squares formula, we can write

This method can be used to remove square roots from an expression. Thus, to simplify a quotient involving square roots, we can multiply the numerator and denominator of the quotient by a conjugate expression first to remove the square roots.

How To: Evaluating Limits of Quotients Involving Square Roots

Let be a quotient of functions involving square roots. To evaluate the limit of in the indeterminate form , we need to

multiply the numerator and denominator of the quotient by a conjugate expression,
simplify and cancel out all common factors,
equate the limit of the simplified quotient to the original limit,
evaluate the limit.

In the next example, we will use this method to evaluate a limit of a function involving square roots.

Example 5: Finding the Limit of a Difference of Powers Involving Roots

Determine .

Answer

In this example, we need to find the limit of a function that is a result of the sum, difference, quotient, and composition of the square root function and polynomial functions. We know that we can find the limit of such functions by direct substitution if the limit point belongs to the domain of the function. For a number to belong to the domain of such a function, we must make sure that the denominator of the function does not equal zero at this number and that the expression under the square root is nonnegative at this number. In short, if we are able to compute the value of the function at the given number, then the number belongs to the domain.

Let us begin by examining what happens to the given function when we substitute the limit point :

The expression is called an indeterminate form. We know that an indeterminate form is never a valid answer to a limit problem. Recall that we can simplify a quotient expression involving square roots by first multiplying the numerator and denominator of the quotient by a conjugate expression. In the given function, the square root expression is in the numerator: . For an expression of the form , the conjugate expression is . We know that multiplying by conjugates can remove square roots by

We can write which leads to the conjugate expression . We can multiply the top and the bottom of this quotient by this expression to obtain

We can see that the denominator of the resulting function does not equal zero at the limit point , so we can evaluate this limit by direct substitution:

This leads to

In the next example, we will consider the limit of a quotient where both the numerator and denominator contain square root expressions.

Example 6: Finding the Limit of a Combination of Root Functions at a Point Using Rationalization

Find .

Answer

In this example, we need to find the limit of a function that is a result of the difference, quotient, and composition of the square root function and polynomial functions. We know that we can find the limit of such a function by direct substitution if the limit point belongs to the domain of the function. For a number to belong to the domain of such a function, we must make sure that the denominator of the function does not equal zero at this number and that the expression under the square root is nonnegative at this number. In short, if we are able to compute the value of the function at the given number, then the number belongs to the domain.

Let us begin by examining what happens to the given function when we substitute the limit point :

The expression is called an indeterminate form. We know that an indeterminate form is never a valid answer to a limit problem. Recall that we can simplify a quotient expression involving square roots by first multiplying the numerator and denominator of the quotient by a conjugate expression. In the given quotient, both the numerator and denominator contain two square root expressions. For an expression of the form , the conjugate expression is . We know that multiplying by conjugates can remove square roots by

This means that the conjugate of is and that the conjugate of is . We need to multiply the numerator and denominator of this quotient by both of these factors:

We can see that the denominator of the resulting function does not equal zero at the limit point , so we can evaluate this limit by direct substitution:

This leads to

Another type of quotient in indeterminate form contains difference in quotients. If the limit in an indeterminate form contains the difference or sum of quotients, then we need to first simplify the expression by finding the common denominator.

How To: Evaluating Limits Involving Difference of Quotients

Let be a function whose expression involves the sum or difference of quotients. If the limit of assumes an indeterminate form, we need to

find the sum or difference of the quotients by finding the common denominator,
simplify and cancel out all common factors,
equate the limit of the simplified quotient to the original limit,
evaluate the limit.

In our final example, we will find the limit of a function involving difference of quotients assuming an indeterminate form.

Example 7: Finding the Limit of a Combination of Rational and Polynomial Functions at a Point

Find .

Answer

In this example, we need to find the limit of a quotient whose numerator is a difference of smaller quotients. The function we are taking the limit of is the result of a difference and quotient of a rational function, a constant, and a polynomial. We know that we can find the limit of a rational function by direct substitution only if the limit point belongs to the domain of the function. We know that a number belongs to the domain of a function if we are able to compute the value of the function at the number.

Let us begin by examining what happens to the given function when we substitute the limit point :

The expression is called an indeterminate form. We know that an indeterminate form is never a valid answer to a limit problem. Recall that we can evaluate the limit involving the sum or difference of quotients by finding the sum or difference. To find the difference of two quotients, we need to find the common denominator. The common denominator between and 6 is . Hence, the difference of quotients in the numerator of the given function is

This simplifies the numerator of the given quotient. The denominator of the given quotient is , which is the same as multiplying the numerator by . Hence,

Since the denominator of the rational function in the limit of the right-hand side of this equation does not equal zero, we can evaluate this limit by direct substitution:

Hence,

Let us finish by recapping a few important concepts from this explainer.

Key Points

A limit of the form is said to be in the indeterminate form if both the numerator and denominator of the quotient approach zero at . The indeterminate form is never a valid answer to a limit problem, but it means that we need to work harder to evaluate this limit.
Using algebraic methods to evaluate a limit means that we first simplify the given expression to the point where we can evaluate the limit by direct substitution. We can do this using the property that if for all , then
Simplifying functions to evaluate the limits requires different steps for different types of functions:
- For rational functions, we can factor both the numerator and denominator of the quotient and then cancel out the common factor to simplify.
- For a quotient involving square root expressions, we need to first multiply the numerator and denominator of the quotient by a conjugate expression before simplifying.
- For an expression involving the sum or difference of quotients, we need to first find the sum or difference of the quotients using the common denominator.
When the limit of a rational function at is in an indeterminate form , both the numerator and denominator must contain the factor . If the factorization is difficult, we can use the polynomial long division by the factor to factor each polynomial in the form for some other polynomial . We can then cancel this factor out and find the limit by direct substitution.