Lesson Explainer: Limits of a Difference of Powers | Nagwa Lesson Explainer: Limits of a Difference of Powers | Nagwa

Lesson Explainer: Limits of a Difference of Powers Mathematics

In this explainer, we will learn how to evaluate limits of a difference of powers.

Before we discuss how to evaluate the limit of a difference of powers of π‘₯, let’s briefly recall the definition of a limit and some of the properties we will need.

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.

Following this definition, it is possible to show the following properties of the limits of functions at a point.

Properties: Limits of functions

If limο—β†’οŒΊοŠ§π‘“(π‘₯)=𝐿, limο—β†’οŒΊοŠ¨π‘”(π‘₯)=𝐿, and π‘˜ is any constant, then

  • limο—β†’οŒΊοŠ§(π‘˜π‘“(π‘₯))=π‘˜πΏ,
  • limο—β†’οŒΊοŠ§οŠ¨(𝑓(π‘₯)±𝑔(π‘₯))=𝐿±𝐿,
  • limο—β†’οŒΊοŠ§οŠ¨(𝑓(π‘₯)×𝑔(π‘₯))=𝐿×𝐿,
  • limο—β†’οŒΊοŠ§οŠ¨ο€½π‘“(π‘₯)𝑔(π‘₯)=𝐿𝐿, if 𝐿≠0.

If π‘›βˆˆβ„ and (𝐿)βˆˆβ„οŠ§οŠ, then limο—β†’οŒΊοŠοŠ§οŠ(𝑓(π‘₯))=(𝐿).

If β„Ž(π‘₯) is a polynomial, then, for any π‘Žβˆˆβ„, limο—β†’οŒΊβ„Ž(π‘₯)=β„Ž(π‘Ž).

Using the properties of limits, we can evaluate any rational function. Since a rational function is the quotient of two polynomials, say, 𝑃(π‘₯)𝑄(π‘₯), then if 𝑄(π‘Ž)β‰ 0, we have limο—β†’οŒΊο€½π‘ƒ(π‘₯)𝑄(π‘₯)=𝑃(π‘Ž)𝑄(π‘Ž).

We want to extend this to the case when 𝑄(π‘Ž)=0. Let’s start by considering an example. Let 𝑓(π‘₯)=(π‘₯βˆ’1)(π‘₯+1)π‘₯+1. To determine the limit of 𝑓(π‘₯) as π‘₯ approaches βˆ’1, we can sketch its graph.

We can see from the graph that as the values of π‘₯ approach βˆ’1 from either side, the outputs approach βˆ’2.

Recall that the value of 𝑓(βˆ’1) does not affect its limit as π‘₯ approaches βˆ’1. This means we can choose any value we want for 𝑓(βˆ’1) and still evaluate its limit. In particular, if we were to say 𝑓(βˆ’1)=βˆ’2, we would get the following graph.

This is then the line 𝑦=π‘₯βˆ’1. We can evaluate the limit of this function by direct substitution and this is the same as the limit of 𝑓(π‘₯): limlimο—β†’οŠ±οŠ§ο—β†’οŠ±οŠ§π‘“(π‘₯)=π‘₯βˆ’1=βˆ’1βˆ’1=βˆ’2.

We can formalize this result directly from the definition of a limit. Since 𝑓(π‘Ž) does not affect the limit of 𝑓 as π‘₯ approaches π‘Ž, any two functions that agree everywhere except at π‘₯=π‘Ž must have the same limit at π‘Ž.

Property: Functions with the Same Limit

If 𝑓(π‘₯)=𝑔(π‘₯) for all π‘₯βˆˆβ„βˆ’{π‘Ž} and limο—β†’οŒΊπ‘”(π‘₯)=𝐿, then limο—β†’οŒΊπ‘“(π‘₯)=𝐿.

Therefore, if we are evaluating the limit of a function, we can manipulate the function by changing its value at the limit point. In the above example, we used the fact that when π‘₯β‰ βˆ’1, π‘₯+1 is not zero, so 𝑓(π‘₯)=(π‘₯+1)(π‘₯βˆ’1)π‘₯+1=(π‘₯+1)(π‘₯βˆ’1)π‘₯+1=π‘₯βˆ’1π‘₯β‰ βˆ’1.if

In other words, if 𝑔(π‘₯)=π‘₯βˆ’1, then 𝑓(π‘₯)=𝑔(π‘₯) for all π‘₯βˆˆβ„βˆ’{βˆ’1}. Hence, limlimlimο—β†’οŠ±οŠ§ο—β†’οŠ±οŠ§ο—β†’οŠ±οŠ§π‘“(π‘₯)=𝑔(π‘₯)=π‘₯βˆ’1=βˆ’2.

We can evaluate the limit of a rational function by canceling shared factors; since this will not change the value of the function around the limit point, it can only change the value of the function at that point.

Let’s see an example of applying this to evaluate the limit of a rational function.

Example 1: Finding the Limit of a Function Using Difference of Powers

Find limο—β†’οŠ«οŠͺπ‘₯βˆ’625π‘₯βˆ’125.

Answer

Since we are asked to find the limit of a rational function, we can start by trying direct substitution: 5βˆ’6255βˆ’125=00.οŠͺ

Since this evaluates to give us an indeterminate form, we cannot conclude anything about this limit by direct substitution. Instead, let’s fully factor the numerator and denominator of the rational function. It might help to note that we know both have a factor of π‘₯βˆ’5 by the remainder theorem: π‘₯βˆ’625=ο€Ήπ‘₯βˆ’25π‘₯+25=(π‘₯βˆ’5)(π‘₯+5)ο€Ήπ‘₯+25,π‘₯βˆ’125=(π‘₯βˆ’5)ο€Ήπ‘₯+5π‘₯+25.οŠͺ

Hence, limlimο—β†’οŠ«οŠͺοŠ©ο—β†’οŠ«οŠ¨οŠ¨π‘₯βˆ’625π‘₯βˆ’125=(π‘₯βˆ’5)(π‘₯+5)ο€Ήπ‘₯+25(π‘₯βˆ’5)(π‘₯+5π‘₯+25).

Recall that if 𝑓(π‘₯)=𝑔(π‘₯) for all π‘₯βˆˆβ„βˆ’{π‘Ž} and limο—β†’οŒΊπ‘”(π‘₯)=𝐿, then limο—β†’οŒΊπ‘“(π‘₯)=𝐿. This allows us to cancel the shared factor of π‘₯βˆ’5 in the numerator and denominator, since it will not affect the value of the function anywhere except when π‘₯=5: limlimο—β†’οŠ«οŠ¨οŠ¨ο—β†’οŠ«οŠ¨οŠ¨(π‘₯βˆ’5)(π‘₯+5)ο€Ήπ‘₯+25(π‘₯βˆ’5)(π‘₯+5π‘₯+25)=(π‘₯+5)ο€Ήπ‘₯+25(π‘₯+5π‘₯+25).

We can then evaluate this limit by direct substitution: limο—β†’οŠ«οŠ¨οŠ¨οŠ¨οŠ¨(π‘₯+5)ο€Ήπ‘₯+25(π‘₯+5π‘₯+25)=(5+5)ο€Ή5+25(5+5(5)+25)=50075=203.

Therefore, limο—β†’οŠ«οŠͺπ‘₯βˆ’625π‘₯βˆ’125=203.

We can apply the result above to show some other useful results. For example, if we want to evaluate limο—β†’οŒΊοŠοŠπ‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž, where π‘›βˆˆβ„•, then by the remainder theorem, both the numerator and denominator share a factor of π‘₯βˆ’π‘Ž. We can factor the numerator to evaluate the limit: limlimlimο—β†’οŒΊοŠοŠο—β†’οŒΊοŠοŠ±οŠ§οŠοŠ±οŠ¨οŠ¨οŠοŠ±οŠ©οŠοŠ±οŠ§ο—β†’οŒΊοŠοŠ±οŠ§οŠοŠ±οŠ¨οŠ¨οŠοŠ±οŠ©οŠοŠ±οŠ§οŠοŠ±οŠ§οŠοŠ±οŠ¨οŠ¨οŠοŠ±οŠ©οŠοŠ±οŠ§οŠοŠ±οŠ§π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=(π‘₯βˆ’π‘Ž)ο€Ήπ‘₯+π‘Žπ‘₯+π‘Žπ‘₯+β‹―+π‘Žο…π‘₯βˆ’π‘Ž=ο€Ήπ‘₯+π‘Žπ‘₯+π‘Žπ‘₯+β‹―+π‘Žο…=ο€Ήπ‘Ž+π‘Žο€Ήπ‘Žο…+π‘Žο€Ήπ‘Žο…+β‹―+π‘Žο…=π‘›π‘Ž.

The result we have just shown is true for more than just natural numbers 𝑛.

Theorem: Limit of a Rational Function

For any real constants 𝑛 and π‘Ž, limο—β†’οŒΊοŠοŠοŠοŠ±οŠ§π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘Ž, provided π‘ŽοŠ and π‘ŽοŠοŠ±οŠ§ exist.

We can use this to prove another useful result. We want to evaluate limο—β†’οŒΊοŠοŠο‰ο‰π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž. We can do this by rewriting our limit: limlimο—β†’οŒΊοŠοŠο‰ο‰ο—β†’οŒΊοŠοŠο‰ο‰π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=ο€½π‘₯βˆ’π‘Žπ‘₯βˆ’π‘ŽΓ·π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Žο‰.

Now, we can apply the quotient rule for limits: limlimlimο—β†’οŒΊοŠοŠο‰ο‰ο—β†’οŒΊοŠοŠο—β†’οŒΊο‰ο‰ο€½π‘₯βˆ’π‘Žπ‘₯βˆ’π‘ŽΓ·π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Žο‰=ο€½π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Žο‰Γ·ο€½π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Žο‰.

Provided both limits exist, we can then evaluate these limits using our limit result above: limlimο—β†’οŒΊοŠοŠο—β†’οŒΊο‰ο‰οŠοŠ±οŠ§ο‰οŠ±οŠ§οŠοŠ±ο‰ο€½π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Žο‰Γ·ο€½π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Žο‰=π‘›π‘Žπ‘šπ‘Ž=π‘›π‘šπ‘Ž.

This gives us the following result.

Theorem: Limit of a Difference of Powers

For any 𝑛,π‘š,π‘Žβˆˆβ„, limο—β†’οŒΊοŠοŠο‰ο‰οŠοŠ±ο‰π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘šπ‘Ž, provided π‘šβ‰ 0 and π‘ŽοŠ, π‘Žο‰, and π‘ŽοŠοŠ±ο‰ all exist.

We can use this to evaluate the limit in our first example directly, limο—β†’οŠ«οŠͺπ‘₯βˆ’625π‘₯βˆ’125. In this case, we rewrite the limit as limlimο—β†’οŠ«οŠͺοŠ©ο—β†’οŠ«οŠͺοŠͺπ‘₯βˆ’625π‘₯βˆ’125=π‘₯βˆ’5π‘₯βˆ’5.

We see that 𝑛=4, π‘Ž=5, and π‘š=3. Substituting these values into the difference of powers limit result gives us limο—β†’οŠ«οŠͺοŠͺοŠͺπ‘₯βˆ’5π‘₯βˆ’5=43(5)=203.

There is one more type of limit we can evaluate from this limit result. If we substitute 𝑦=π‘₯βˆ’π‘Ž into limο—β†’οŒΊοŠοŠοŠοŠ±οŠ§π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘Ž, we get π‘›π‘Ž=π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=(𝑦+π‘Ž)βˆ’π‘Žπ‘¦.οŠοŠ±οŠ§ο—β†’οŒΊοŠοŠο˜β†’οŠ¦οŠοŠlimlim

This gives us another result.

Theorem: Limit of a Rational Function

For any real constants 𝑛 and π‘Ž, limο—β†’οŠ¦οŠοŠοŠοŠ±οŠ§(π‘₯+π‘Ž)βˆ’π‘Žπ‘₯=π‘›π‘Ž, provided π‘ŽοŠ and π‘ŽοŠοŠ±οŠ§ exist.

Let’s see some examples of applying these limit results to evaluate the limits of different functions.

Example 2: Finding the Limit of a Difference of Powers Using Algebraic Manipulation

Find limο—β†’οŠ§οŽ₯√π‘₯+√π‘₯βˆ’2π‘₯βˆ’1.

Answer

Since this is the limit of a radical expression, we could attempt to evaluate this limit by direct substitution: οŽ₯√1+√1βˆ’21βˆ’1=00.

Since this gives an indeterminate form, we cannot evaluate this limit by direct substitution. Instead, we recall that, for any real constants 𝑛 and π‘Ž, limο—β†’οŒΊοŠοŠοŠοŠ±οŠ§π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘Ž, provided π‘ŽοŠ and π‘ŽοŠοŠ±οŠ§ exist.

We can rewrite our limit in this form by splitting the fraction over the numerator as follows: limlimlimο—β†’οŠ§ο—β†’οŠ§ο—β†’οŠ§οŽ₯οŽ₯οŽ₯√π‘₯+√π‘₯βˆ’2π‘₯βˆ’1=√π‘₯βˆ’1+√π‘₯βˆ’1π‘₯βˆ’1=ο€Ώβˆšπ‘₯βˆ’1π‘₯βˆ’1+√π‘₯βˆ’1π‘₯βˆ’1.

Then, by using the fact that the limit of a sum is equal to the sum of a limit, provided both limits exist, we have limlimlimο—β†’οŠ§ο—β†’οŠ§ο—β†’οŠ§οŠ±οŠ§οŠ±οŠ§ο€Ώβˆšπ‘₯βˆ’1π‘₯βˆ’1+√π‘₯βˆ’1π‘₯βˆ’1=ο€Ώβˆšπ‘₯βˆ’1π‘₯βˆ’1+ο€Ώβˆšπ‘₯βˆ’1π‘₯βˆ’1=16(1)+122(1)=16+122=733.οŽ₯οŽ₯οŽ₯

Hence, limο—β†’οŠ§οŽ₯√π‘₯+√π‘₯βˆ’2π‘₯βˆ’1=733.

Example 3: Finding the Limit of a Function Involving Roots

Find limο—β†’οŠ§οŠ­οŠ¨ο€Ίβˆšπ‘₯βˆ’1ο†ο€»βˆšπ‘₯βˆ’1(π‘₯βˆ’1)οŽ₯.

Answer

Since this is the limit of a radical expression, we could attempt to evaluate this limit by direct substitution: ο€»βˆš1βˆ’1(1)βˆ’1(1βˆ’1)=00.οŽ₯

Since this gives an indeterminate form, we cannot evaluate this limit by direct substitution. Instead, we recall that, for any real constants 𝑛 and π‘Ž.

limο—β†’οŒΊοŠοŠοŠοŠ±οŠ§π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘Ž, provided π‘ŽοŠ and π‘ŽοŠοŠ±οŠ§ exist.

We can rewrite our limit in this form by using the product rule for limits: limlimlimlimlimlimο—β†’οŠ§οŠ­οŠ¨ο—β†’οŠ§οŠ­ο—β†’οŠ§ο—β†’οŠ§οŠ­ο—β†’οŠ§ο—β†’οŠ§οŠ±οŠ§οŠ±οŠ§ο€Ίβˆšπ‘₯βˆ’1ο†ο€»βˆšπ‘₯βˆ’1(π‘₯βˆ’1)=ο€βˆšπ‘₯βˆ’1π‘₯βˆ’1Γ—βˆšπ‘₯βˆ’1π‘₯βˆ’1=ο€Ώβˆšπ‘₯βˆ’1π‘₯βˆ’1ο‹Γ—ο€βˆšπ‘₯βˆ’1π‘₯βˆ’1=π‘₯βˆ’1π‘₯βˆ’1×π‘₯βˆ’1π‘₯βˆ’1=ο€½14[1]×76[1]=14Γ—76=724.οŽ₯οŽ₯οŽ₯οŽ₯οŽ₯

Hence, limο—β†’οŠ§οŠ­οŠ¨ο€Ίβˆšπ‘₯βˆ’1ο†ο€»βˆšπ‘₯βˆ’1(π‘₯βˆ’1)=724οŽ₯.

In our next example, we will use a substitution to rewrite a limit into a form that we can evaluate.

Example 4: Finding the Limit of a Rational Function

Find limο—β†’οŠ¨οŠ©(π‘₯βˆ’4)+8π‘₯βˆ’2.

Answer

Since this is the limit of a rational function, we can attempt to evaluate the limit by direct substitution: (2βˆ’4)+82βˆ’2=00.

This is an indeterminate form, so we cannot evaluate the limit by direct substitution. To evaluate this limit, we need to notice this limit is similar to one of our limit results involving the limit of a difference of powers. Recall that, for any real constants 𝑛 and π‘Ž, limο—β†’οŠ¦οŠοŠοŠοŠ±οŠ§(π‘₯+π‘Ž)βˆ’π‘Žπ‘₯=π‘›π‘Ž, provided π‘ŽοŠ and π‘ŽοŠοŠ±οŠ§ exist.

To write our limit in this form, we need to use the substitution 𝑦=π‘₯βˆ’2: limlimο—β†’οŠ¨οŠ©ο˜β†’οŠ¦οŠ©(π‘₯βˆ’4)+8π‘₯βˆ’2=(π‘¦βˆ’2)+8𝑦.

This is now in the required form with π‘₯=𝑦, π‘Ž=βˆ’2, and 𝑛=3; hence, limο˜β†’οŠ¦οŠ©οŠ©οŠ±οŠ§(π‘¦βˆ’2)+8𝑦=3(βˆ’2)=12.

Therefore, limο—β†’οŠ¨οŠ©(π‘₯βˆ’4)+8π‘₯βˆ’2=12.

In our final example, we will need to use several algebraic manipulation techniques and limit properties to evaluate the limit of a rational function.

Example 5: Finding the Limit of a Rational Function

Find limο—β†’οŠ§οŠͺοŠͺοŠ©οŠ©οŠ¬ο˜ο€Ήπ‘₯βˆ’1(π‘₯βˆ’1)Γ—1π‘₯βˆ’1.

Answer

Since this is the limit of a rational function, we can attempt to evaluate the limit by direct substitution: ο€Ή1βˆ’1(1βˆ’1)Γ—11βˆ’1=00.οŠͺοŠͺ

This is an indeterminate form, so we cannot evaluate the limit by direct substitution.

To evaluate this limit, we will start by rearranging the rational function by using difference between squares: limlimlimο—β†’οŠ§οŠͺοŠͺοŠ©οŠ©οŠ¬ο—β†’οŠ§οŠͺοŠͺοŠ©οŠ©οŠ©οŠ©ο—β†’οŠ§οŠͺοŠͺοŠͺοŠ©ο˜ο€Ήπ‘₯βˆ’1(π‘₯βˆ’1)Γ—1π‘₯βˆ’1=ο˜ο€Ήπ‘₯βˆ’1(π‘₯βˆ’1)Γ—1(π‘₯βˆ’1)(π‘₯+1)=ο˜ο€Ήπ‘₯βˆ’1(π‘₯βˆ’1)Γ—1π‘₯+1.

We can then evaluate this limit by using the product rule for limits, power rule for limits, and the fact that, for any 𝑛,π‘š,π‘Žβˆˆβ„, limο—β†’οŒΊοŠοŠο‰ο‰οŠοŠ±ο‰π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘šπ‘Ž, provided π‘šβ‰ 0 and π‘ŽοŠ, π‘Žο‰, and π‘ŽοŠοŠ±ο‰ all exist: limlimlimlimο—β†’οŠ§οŠͺοŠͺοŠͺοŠ©ο—β†’οŠ§οŠͺοŠͺοŠͺο—β†’οŠ§οŠ©ο—β†’οŠ§οŠͺοŠͺοŠͺοŠͺο˜ο€Ήπ‘₯βˆ’1(π‘₯βˆ’1)Γ—1π‘₯+1=ο˜ο€Ήπ‘₯βˆ’1(π‘₯βˆ’1)×1π‘₯+1=π‘₯βˆ’1(π‘₯βˆ’1)×11+1=43(1)×12=12881.

Hence, limο—β†’οŠ§οŠͺοŠͺοŠ©οŠ©οŠ¬ο˜ο€Ήπ‘₯βˆ’1(π‘₯βˆ’1)Γ—1π‘₯βˆ’1=12881.

Let’s finish by recapping some of the important points of this explainer.

Key Points

  • If 𝑓(π‘₯)=𝑔(π‘₯) for all π‘₯βˆˆβ„βˆ’{π‘Ž} and limο—β†’οŒΊπ‘”(π‘₯)=𝐿, then limο—β†’οŒΊπ‘“(π‘₯)=𝐿.
  • For any real constants 𝑛 and π‘Ž, limprovidedandexistο—β†’οŒΊοŠοŠοŠοŠ±οŠ§οŠοŠοŠ±οŠ§π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘Ž,π‘Žπ‘Ž.
  • For any 𝑛,π‘š,π‘Žβˆˆβ„, limprovidedandandallexistο—β†’οŒΊοŠοŠο‰ο‰οŠοŠ±ο‰οŠο‰οŠοŠ±ο‰π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘šπ‘Ž,π‘šβ‰ 0π‘Ž,π‘Ž,π‘Ž.
  • For any real constants 𝑛 and π‘Ž, limprovidedandexistο—β†’οŠ¦οŠοŠοŠοŠ±οŠ§οŠοŠοŠ±οŠ§(π‘₯+π‘Ž)βˆ’π‘Žπ‘₯=π‘›π‘Ž,π‘Žπ‘Ž.

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