Lesson Explainer: Limits by Direct Substitution Mathematics • Higher Education

In this explainer, we will learn how to use the direct substitution method to evaluate limits.

The limit of a function at a point represents the behavior of the function near that point, rather than the value of the function at the point. Let us begin by recalling the formal definition of the limit of a function.

Definition: Limit of a Function

If the values of 𝑓(π‘₯) approach some value of 𝐿 as the values of π‘₯ approach π‘Ž (from both sides), but not necessarily when π‘₯=π‘Ž, then we say the limit of 𝑓(π‘₯) as π‘₯ approaches π‘Ž is equal to 𝐿 and we denote this as limο—β†’οŒΊπ‘“(π‘₯)=𝐿.

From this definition of the limit, we can immediately understand the limit of a constant function.

Definition: Limit of Constant Functions

Let 𝐢 be a constant. Then, for any real number π‘Ž, we have limο—β†’οŒΊπΆ=𝐢.

This is because the constant function 𝐢 always takes the value 𝐢, regardless of the input π‘₯. In fact, we can observe that a constant function does not have an β€œπ‘₯” to substitute the input value into. Since the value of a constant function is always 𝐢, we can say that the function approaches 𝐢 as π‘₯ approaches π‘Ž, regardless of the value of π‘Ž.

In our first example, we will find the limit of a constant function.

Example 1: Finding the Limit of a Constant Function

Find limο—β†’οŠ±οŠ§(30).

Answer

In this example, we need to find the limit of a constant function. Recall that, for any constant 𝐢 and a limit point π‘Ž, limο—β†’οŒΊπΆ=𝐢.

In other words, the limit of a constant function is always equal to the constant value, regardless of the location of the limit point. In this example, the constant value is 30. Hence, limο—β†’οŠ±οŠ§(30)=30.

In the previous example, we found the limit of a constant function. Let us turn our attention to more interesting functions.

One way that we can find the limit of a function is by using a table containing the value of the function near a point. For instance, say that the value of function 𝑓(π‘₯) near π‘₯=0 is given in the table below.

π‘₯βˆ’1βˆ’0.5βˆ’0.1βˆ’0.010.010.10.51
𝑓(π‘₯)00.50.90.990.990.90.40

From the table above, we can observe that the value of 𝑓(π‘₯) approaches 1 as π‘₯ approaches 0 from either side. This leads to the fact that the limit of 𝑓(π‘₯) as π‘₯ approaches 0 is equal to 1, or equivalently limο—β†’οŠ¦π‘“(π‘₯)=1.

However, the table does not include the value of 𝑓(π‘₯) at π‘₯=0. Based on this limit, it is reasonable to suspect that 𝑓(0)=1 since the pattern of the behavior of this function appears to lead to such a conclusion. On one hand, it is important to keep in mind that the limit value does not have to equal the function value. Just because the function behaves in such a way near a point, it does not have to produce that value at that point. It is possible that this function 𝑓(π‘₯) has a different value at π‘₯=0, for instance 𝑓(0)=10.

On the other hand, most of the functions familiar to us do not behave this way. Loosely speaking, if this function 𝑓(π‘₯) is a β€œtypical” function, then we expect it to satisfy 𝑓(0)=1. This means that for a large family of functions, we can find the limit of the function by computing the function value at that point. This method of evaluating the limit of a function is called the direct substitution method.

In this explainer, we will generate a list of functions from which we expect this behavior. We should keep in mind that this method is not valid for all functions, but just for the β€œtypical” ones, which is a loosely defined term. Therefore, it is useful to know precisely what types of functions are eligible for the direct substitution method.

Rule: Limit of Polynomial Functions by Direct Substitution

Let 𝑓(π‘₯) be a polynomial. Then, for any π‘Žβˆˆβ„, limο—β†’οŒΊπ‘“(π‘₯)=𝑓(π‘Ž).

It is worth noting that a constant function is a special case of polynomial functions, which means that we can also apply the direct substitution method for constant functions. Since applying the direct substitution to a constant function always produces the same constant value, we can say that limο—β†’οŒΊπΆ=𝐢, for any real numbers π‘Ž and 𝐢. This recovers the property that we stated earlier.

We can understand the direct substitution rule better when we consider the graph of a polynomial function. The graph of a polynomial function is always a smooth and connected curve. For instance, consider the graph of a particular polynomial below.

From the graph of this polynomial, we can see the limit of this function at π‘₯=2 by following the graph as π‘₯-value approaches 2 from either side, as indicated by the blue arrows. This tells us that the limit of this function as π‘₯ approaches 2 is the 𝑦-coordinate of the red dot on the graph.

On the other hand, since the red dot is on the graph of the function, we know that the 𝑦-coordinate of this point is the function value at π‘₯=2. Hence, denoting this polynomial function as 𝑓(π‘₯), we can write limο—β†’οŠ¨π‘“(π‘₯)=𝑓(2).

This is the main idea behind evaluating a limit by direct substitution. While we do not usually see the graph of the function in these problems, we are assuming that we will arrive at a point on the graph of the function if we follow the graph of the function. As mentioned earlier, this is usually true. In particular, the statement above is saying that this is always true for a polynomial function.

In our next example, we will find the limit of a polynomial function by direct substitution.

Example 2: Finding the Limit of a Polynomial Function

Determine limο—β†’οŠ±οŠ«οŠ¨ο€Ήβˆ’9π‘₯βˆ’6π‘₯βˆ’9.

Answer

In this example, we need to determine the limit of a polynomial function. Recall that we can use direct substitution to find the limit of a polynomial function. Using the direct substitution method means that we compute the limit of the function by finding the function value. Using limit notations, the direct substitution tells us that, for an eligible function 𝑓(π‘₯), limο—β†’οŒΊπ‘“(π‘₯)=𝑓(π‘Ž).

In this example, 𝑓(π‘₯)=βˆ’9π‘₯βˆ’6π‘₯βˆ’9 and π‘Ž=βˆ’5. We can compute the function value as follows: 𝑓(βˆ’5)=βˆ’9(βˆ’5)βˆ’6(βˆ’5)βˆ’9=βˆ’204.

This function is a polynomial, so it is eligible for the direct substitution method. Hence, limο—β†’οŠ±οŠ«οŠ¨ο€Ήβˆ’9π‘₯βˆ’6π‘₯βˆ’9=βˆ’204.

In the previous example, we computed the limit of a polynomial function by the direct substitution method. We recall that the domain of a polynomial function is the set of all real numbers. In other words, a polynomial function is defined for any input values. This is a convenient feature for the direct substitution method because we can compute the function value at any limit point if our function is a polynomial.

This is not true for many functions that we know. When the limit point π‘Ž is not in the domain of the function 𝑓(π‘₯), we cannot compute 𝑓(π‘Ž), which means that we cannot use the direct substitution method to find the limit of the function. Before we apply the direct substitution method, we need to check that the limit point belongs to the domain of the function.

Fortunately, many functions are still eligible for the direct substitution method despite their domain restrictions. In these cases, we just need to be more cautious of the domain restrictions.

Rule: Limit of Rational Functions by Direct Substitution

Let 𝑓(π‘₯) be a rational function. That is 𝑓(π‘₯)=𝑝(π‘₯)π‘ž(π‘₯) for some polynomials 𝑝(π‘₯) and π‘ž(π‘₯). Then, for any π‘Žβˆˆβ„ such that π‘ž(π‘Ž)β‰ 0, limο—β†’οŒΊπ‘“(π‘₯)=𝑓(π‘Ž).

In the statement above, we note that as long as π‘ž(π‘Ž)β‰ 0, π‘Ž belongs to the domain of the rational function 𝑓(π‘₯). This means that a rational function is eligible for the direct substitution method as long as the limit point belongs to its domain.

In the next example, we will find the value of an unknown in a limit statement involving a rational function.

Example 3: Finding an Unknown in a Rational Function given Its Limit at a Point

Given that limο—β†’οŠ±οŠ«π‘Žπ‘₯βˆ’1=6, what is π‘Ž?

Answer

We are given a limit statement involving a rational function. Recall that we can find the limit of a rational function using the direct substitution method if the limit point belongs to the domain of the function. Using the direct substitution method means that we compute the limit of the function by finding the function value. Using limit notations, the direct substitution tells us that, for an eligible function 𝑓(π‘₯), limο—β†’οŒ»π‘“(π‘₯)=𝑓(𝑏).

In this example, the limit point is π‘₯=βˆ’5 and the rational function is 𝑓(π‘₯)=π‘Žπ‘₯βˆ’1. We recall that the domain of the rational function is the set of real numbers where the denominator is not equal to zero. This means π‘₯βˆ’1β‰ 0, which leads to π‘₯β‰ 1. Since βˆ’5β‰ 1, we know that βˆ’5 belongs to the domain of this function.

We can compute 𝑓(βˆ’5)=π‘Žβˆ’5βˆ’1=βˆ’π‘Ž6.

Thus, using the direct substitution method, we can write limο—β†’οŠ±οŠ«π‘Žπ‘₯βˆ’π‘Ž=βˆ’π‘Ž6.

Since we are already given that this limit is equal to 6, we can set the right-hand side of the equation above to 6: βˆ’π‘Ž6=6.

This leads to π‘Ž=βˆ’36.

If we have functions that are eligible for the direct substitution method, then the sum, difference, product, and quotients of these functions are all eligible for this method, as long as the denominator of the quotient is not equal to zero at the limit point.

Property: Limit of the Sum, Difference, Product, and Quotient of Functions by Direct Substitution

Let 𝑓(π‘₯) and 𝑔(π‘₯) be functions satisfying limlimο—β†’οŒΊο—β†’οŒΊπ‘“(π‘₯)=𝑓(π‘Ž),𝑔(π‘₯)=𝑔(π‘Ž).

Then, limlimlimlimifο—β†’οŒΊο—β†’οŒΊο—β†’οŒΊο—β†’οŒΊ(𝑓(π‘₯)+𝑔(π‘₯))=𝑓(π‘Ž)+𝑔(π‘Ž),(𝑓(π‘₯)βˆ’π‘”(π‘₯))=𝑓(π‘Ž)βˆ’π‘”(π‘Ž),(𝑓(π‘₯)𝑔(π‘₯))=𝑓(π‘Ž)𝑔(π‘Ž),𝑓(π‘₯)𝑔(π‘₯)=𝑓(π‘Ž)𝑔(π‘Ž)𝑔(π‘Ž)β‰ 0.

In our next example, we will use this property to find the limit of a function by direct substitution.

Example 4: Finding the Limit of a Combination of Rational Functions at a Point Using Direct Substitution

Determine limο—β†’οŠ±οŠ«οŠ§ο—οŠ°οŠ­οŠ§οŠ­βˆ’π‘₯.

Answer

In this example, we need to find the limit of a given expression. The simplest way to find the limit is by direct substitution, which means that we are substituting the limit point π‘₯=βˆ’5 into the given expression to obtain the limit value. But we know that not all functions are eligible for the direct substitution method, so let us examine the given function to determine whether this method is valid.

In the numerator of this function, we see a difference of two fractions. The first fraction 1π‘₯+7 is a rational function. We recall that a rational function is eligible for the direct substitution method as long as the limit point belongs to its domain. The domain of this rational function is the set of values where π‘₯β‰ βˆ’7. Since βˆ’5β‰ βˆ’7, it is in the domain, and hence, 1π‘₯+7 is eligible for direct substitution. The second fraction 17 is constant, which can be considered to be a polynomial, which is always eligible for direct substitution.

We also recall that the difference of two functions that are eligible for direct substitution is also eligible for direct substitution. This tells us that the numerator 1π‘₯+7βˆ’17 is eligible for direct substitution.

Finally, we need to consider the denominator π‘₯. We recall that a quotient of two functions that are eligible for direct substitution is also eligible for direct substitution as long as the denominator of the quotient is not equal to zero at the limit point. We can see that the denominator π‘₯ is not equal to zero at π‘₯=βˆ’5; hence, we can apply the direct substitution method to this limit.

Substituting βˆ’5 into the given function, limο—β†’οŠ±οŠ«οŠ§ο—οŠ°οŠ­οŠ§οŠ­οŠ§οŠ±οŠ«οŠ°οŠ­οŠ§οŠ­οŠ§οŠ¨οŠ§οŠ­οŠ­οŠ§οŠͺοŠͺοŠͺβˆ’π‘₯=βˆ’βˆ’5=βˆ’βˆ’5=βˆ’βˆ’5=βˆ’5=514Γ—1βˆ’5=βˆ’114.

Hence, the limit is equal to βˆ’114.

So far, we have found limits of polynomial and rational functions using direct substitution. We can apply this function to a much wider family of functions, as long as the limit point belongs to the domain of the function.

Property: Finding the Limit of a Function by Direct Substitution

The direct substitution method works for the composition of any of the following functions if the limit point belongs to the domain of the resulting function:

  • polynomial or constant functions,
  • rational functions,
  • power or root functions: π‘₯ for some constant 𝑝,
  • trigonometric functions: sinπ‘₯, cosπ‘₯, tanπ‘₯,
  • exponential and log functions: 𝑏 or logπ‘₯ for some 𝑏>0 and 𝑏≠1,
  • absolute value function: |π‘₯|.

Let us consider an example where we find the limit of a square root of a polynomial at a point.

Example 5: Finding the Limit of Root Functions at a Point by Direct Substitution

Determine limο—β†’οŠ―οŠ¨βˆš4π‘₯βˆ’9π‘₯+1.

Answer

In this example, we need to find the limit of a function. We can see that the given function is a composition of a square root and a polynomial function. Recall that we can find the limit of a composition of a polynomial and root functions by direct substitution if the limit point belongs to the domain of the function.

The domain of the given root function is the set of π‘₯-values for which the polynomial under the root gives a nonnegative value. To determine whether the limit point 9 belongs to the domain of the given function, we need to check whether the polynomial under the root is nonnegative when π‘₯=9. We can compute 4(9)βˆ’9Γ—9+1=244.

Since 244 is nonnegative, we can see that 9 belongs to the domain of the given root function. Applying the direct substitution method, limο—β†’οŠ―οŠ¨οŠ¨βˆš4π‘₯βˆ’9π‘₯+1=4(9)βˆ’9Γ—9+1=√244=2√61.

Hence, the given limit is equal to 2√61.

Let us consider another example involving absolute value functions.

Example 6: Finding the Limit of a Function Involving Absolute Value

Given 𝑓(π‘₯)=|π‘₯+11|βˆ’|π‘₯βˆ’18|, find lim→οŠͺ𝑓(π‘₯).

Answer

In this example, we need to find the limit of a function. We can see that the given function is a difference of absolute value functions. Recall that we can find the limit of each absolute value function by direct substitution. We also know that if two functions are eligible for the direct substitution method, their difference is also eligible for direct substitution. Thus, we can find the given limit by direct substitution, which means lim→οŠͺ𝑓(π‘₯)=𝑓(4).

Substituting π‘₯=4 into the function, |4+11|βˆ’|4βˆ’18|=15βˆ’14=1.

Hence, the given limit is equal to 1.

In our final example, we will compute the limit of a function involving rational and trigonometric functions.

Example 7: Finding Limits Involving a Combination of Trigonometric and Quadratic Functions

Find limcosο—β†’οŠ¨οŽ οŽ‘π‘₯(2βˆ’4π‘₯)π‘₯+π‘₯.

Answer

In this example, we need to find the limit of a function. The given function consists of polynomials and a trigonometric function, which are put together using the composition, product, and quotient. We know that the composition, product, and quotient of these polynomials and trigonometric functions are eligible for the direct substitution method as long as the limit point belongs to the domain of the function. Hence, we need to first check whether the limit point 12 belongs to the domain of the given function.

We know that the cosine function and the polynomial functions do not have domain constraints, so the limit point 12 would be in the domain of the given function if the denominator of the quotient is not equal to zero at this point. Let us compute the denominator π‘₯+π‘₯ at π‘₯=12, ο€Ό12+12=14+12=34.

Since the denominator is not equal to zero at this point, 12 is in the domain of the given function. We can then apply the direct substitution method to compute limcoscoscosο—β†’οŠ¨οŠ§οŠ¨οŠ§οŠ¨οŠ§οŠ¨οŠ¨οŠ§οŠ¨οŠ§οŠ¨οŠ©οŠͺπ‘₯(2βˆ’4π‘₯)π‘₯+π‘₯=ο€»2βˆ’4×+=(0).

Using cos0=1, this is equal to οŠͺ=12Γ—43=23.

Thus, the given limit is equal to 23.

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

Key Points

  • When we find the limit of a function by direct substitution, we are substituting the limit point into the given function to find the limit. This means that, for eligible functions, limο—β†’οŒΊπ‘“(π‘₯)=𝑓(π‘Ž).
  • For any constant 𝐢 and a limit point π‘Ž, limο—β†’οŒΊπΆ=𝐢.
  • We can find the limit of any of the sum, difference, product, quotient, and composition of any of the functions listed below using direct substitution, as long as the limit point is in the domain of the given function:
    • polynomial or constant function,
    • rational function,
    • power or root function,
    • exponential or logarithmic function,
    • trigonometric function,
    • absolute value function.

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