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 • Second Year of Secondary School

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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 𝑥, 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(𝑥) approach some value of 𝐿 as the values of 𝑥 approach 𝑎 (from both sides) but not necessarily when 𝑥=𝑎, then we say the limit of 𝑓(𝑥) as 𝑥 approaches 𝑎 is equal to 𝐿 and we denote this as lim𝑓(𝑥)=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 lim𝑓(𝑥)=𝐿, lim𝑔(𝑥)=𝐿, and 𝑘 is any constant, then

  • lim(𝑘𝑓(𝑥))=𝑘𝐿,
  • lim(𝑓(𝑥)±𝑔(𝑥))=𝐿±𝐿,
  • lim(𝑓(𝑥)×𝑔(𝑥))=𝐿×𝐿,
  • lim𝑓(𝑥)𝑔(𝑥)=𝐿𝐿, if 𝐿0.

If 𝑛 and (𝐿), then lim(𝑓(𝑥))=(𝐿).

If (𝑥) is a polynomial, then, for any 𝑎, lim(𝑥)=(𝑎).

Using the properties of limits, we can evaluate any rational function. Since a rational function is the quotient of two polynomials, say, 𝑃(𝑥)𝑄(𝑥), then if 𝑄(𝑎)0, we have lim𝑃(𝑥)𝑄(𝑥)=𝑃(𝑎)𝑄(𝑎).

We want to extend this to the case when 𝑄(𝑎)=0. Let’s start by considering an example. Let 𝑓(𝑥)=(𝑥1)(𝑥+1)𝑥+1. To determine the limit of 𝑓(𝑥) as 𝑥 approaches 1, we can sketch its graph.

We can see from the graph that as the values of 𝑥 approach 1 from either side, the outputs approach 2.

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

This is then the line 𝑦=𝑥1. We can evaluate the limit of this function by direct substitution and this is the same as the limit of 𝑓(𝑥): limlim𝑓(𝑥)=𝑥1=11=2.

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

Property: Functions with the Same Limit

If 𝑓(𝑥)=𝑔(𝑥) for all 𝑥{𝑎} and lim𝑔(𝑥)=𝐿, then lim𝑓(𝑥)=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 𝑥1, 𝑥+1 is not zero, so 𝑓(𝑥)=(𝑥+1)(𝑥1)𝑥+1=(𝑥+1)(𝑥1)𝑥+1=𝑥1𝑥1.if

In other words, if 𝑔(𝑥)=𝑥1, then 𝑓(𝑥)=𝑔(𝑥) for all 𝑥{1}. Hence, limlimlim𝑓(𝑥)=𝑔(𝑥)=𝑥1=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 lim𝑥625𝑥125.

Answer

Since we are asked to find the limit of a rational function, we can start by trying direct substitution: 56255125=00.

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 𝑥5 by the remainder theorem: 𝑥625=𝑥25𝑥+25=(𝑥5)(𝑥+5)𝑥+25,𝑥125=(𝑥5)𝑥+5𝑥+25.

Hence, limlim𝑥625𝑥125=(𝑥5)(𝑥+5)𝑥+25(𝑥5)(𝑥+5𝑥+25).

Recall that if 𝑓(𝑥)=𝑔(𝑥) for all 𝑥{𝑎} and lim𝑔(𝑥)=𝐿, then lim𝑓(𝑥)=𝐿. This allows us to cancel the shared factor of 𝑥5 in the numerator and denominator, since it will not affect the value of the function anywhere except when 𝑥=5: limlim(𝑥5)(𝑥+5)𝑥+25(𝑥5)(𝑥+5𝑥+25)=(𝑥+5)𝑥+25(𝑥+5𝑥+25).

We can then evaluate this limit by direct substitution: lim(𝑥+5)𝑥+25(𝑥+5𝑥+25)=(5+5)5+25(5+5(5)+25)=50075=203.

Therefore, lim𝑥625𝑥125=203.

We can apply the result above to show some other useful results. For example, if we want to evaluate lim𝑥𝑎𝑥𝑎, where 𝑛, then by the remainder theorem, both the numerator and denominator share a factor of 𝑥𝑎. We can factor the numerator to evaluate the limit: limlimlim𝑥𝑎𝑥𝑎=(𝑥𝑎)𝑥+𝑎𝑥+𝑎𝑥++𝑎𝑥𝑎=𝑥+𝑎𝑥+𝑎𝑥++𝑎=𝑎+𝑎𝑎+𝑎𝑎++𝑎=𝑛𝑎.

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 𝑎, lim𝑥𝑎𝑥𝑎=𝑛𝑎, provided 𝑎 and 𝑎 exist.

We can use this to prove another useful result. We want to evaluate lim𝑥𝑎𝑥𝑎. We can do this by rewriting our limit: limlim𝑥𝑎𝑥𝑎=𝑥𝑎𝑥𝑎÷𝑥𝑎𝑥𝑎.

Now, we can apply the quotient rule for limits: limlimlim𝑥𝑎𝑥𝑎÷𝑥𝑎𝑥𝑎=𝑥𝑎𝑥𝑎÷𝑥𝑎𝑥𝑎.

Provided both limits exist, we can then evaluate these limits using our limit result above: limlim𝑥𝑎𝑥𝑎÷𝑥𝑎𝑥𝑎=𝑛𝑎𝑚𝑎=𝑛𝑚𝑎.

This gives us the following result.

Theorem: Limit of a Difference of Powers

For any 𝑛,𝑚,𝑎, lim𝑥𝑎𝑥𝑎=𝑛𝑚𝑎, provided 𝑚0 and 𝑎, 𝑎, and 𝑎 all exist.

We can use this to evaluate the limit in our first example directly, lim𝑥625𝑥125. In this case, we rewrite the limit as limlim𝑥625𝑥125=𝑥5𝑥5.

We see that 𝑛=4, 𝑎=5, and 𝑚=3. Substituting these values into the difference of powers limit result gives us lim𝑥5𝑥5=43(5)=203.

There is one more type of limit we can evaluate from this limit result. If we substitute 𝑦=𝑥𝑎 into lim𝑥𝑎𝑥𝑎=𝑛𝑎, we get 𝑛𝑎=𝑥𝑎𝑥𝑎=(𝑦+𝑎)𝑎𝑦.limlim

This gives us another result.

Theorem: Limit of a Rational Function

For any real constants 𝑛 and 𝑎, lim(𝑥+𝑎)𝑎𝑥=𝑛𝑎, provided 𝑎 and 𝑎 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 lim𝑥+𝑥2𝑥1.

Answer

Since this is the limit of a radical expression, we could attempt to evaluate this limit by direct substitution: 1+1211=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 𝑎, lim𝑥𝑎𝑥𝑎=𝑛𝑎, provided 𝑎 and 𝑎 exist.

We can rewrite our limit in this form by splitting the fraction over the numerator as follows: limlimlim𝑥+𝑥2𝑥1=𝑥1+𝑥1𝑥1=𝑥1𝑥1+𝑥1𝑥1.

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 limlimlim𝑥1𝑥1+𝑥1𝑥1=𝑥1𝑥1+𝑥1𝑥1=16(1)+122(1)=16+122=733.

Hence, lim𝑥+𝑥2𝑥1=733.

Example 3: Finding the Limit of a Function Involving Roots

Find lim𝑥1𝑥1(𝑥1).

Answer

Since this is the limit of a radical expression, we could attempt to evaluate this limit by direct substitution: 11(1)1(11)=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 𝑎.

lim𝑥𝑎𝑥𝑎=𝑛𝑎, provided 𝑎 and 𝑎 exist.

We can rewrite our limit in this form by using the product rule for limits: limlimlimlimlimlim𝑥1𝑥1(𝑥1)=𝑥1𝑥1×𝑥1𝑥1=𝑥1𝑥1×𝑥1𝑥1=𝑥1𝑥1×𝑥1𝑥1=14[1]×76[1]=14×76=724.

Hence, lim𝑥1𝑥1(𝑥1)=724.

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 lim(𝑥4)+8𝑥2.

Answer

Since this is the limit of a rational function, we can attempt to evaluate the limit by direct substitution: (24)+822=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 𝑎, lim(𝑥+𝑎)𝑎𝑥=𝑛𝑎, provided 𝑎 and 𝑎 exist.

To write our limit in this form, we need to use the substitution 𝑦=𝑥2: limlim(𝑥4)+8𝑥2=(𝑦2)+8𝑦.

This is now in the required form with 𝑥=𝑦, 𝑎=2, and 𝑛=3; hence, lim(𝑦2)+8𝑦=3(2)=12.

Therefore, lim(𝑥4)+8𝑥2=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 lim𝑥1(𝑥1)×1𝑥1.

Answer

Since this is the limit of a rational function, we can attempt to evaluate the limit by direct substitution: 11(11)×111=00.

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: limlimlim𝑥1(𝑥1)×1𝑥1=𝑥1(𝑥1)×1(𝑥1)(𝑥+1)=𝑥1(𝑥1)×1𝑥+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,𝑚,𝑎, lim𝑥𝑎𝑥𝑎=𝑛𝑚𝑎, provided 𝑚0 and 𝑎, 𝑎, and 𝑎 all exist: limlimlimlim𝑥1(𝑥1)×1𝑥+1=𝑥1(𝑥1)×1𝑥+1=𝑥1(𝑥1)×11+1=43(1)×12=12881.

Hence, lim𝑥1(𝑥1)×1𝑥1=12881.

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

Key Points

  • If 𝑓(𝑥)=𝑔(𝑥) for all 𝑥{𝑎} and lim𝑔(𝑥)=𝐿, then lim𝑓(𝑥)=𝐿.
  • For any real constants 𝑛 and 𝑎, limprovidedandexist𝑥𝑎𝑥𝑎=𝑛𝑎,𝑎𝑎.
  • For any 𝑛,𝑚,𝑎, limprovidedandandallexist𝑥𝑎𝑥𝑎=𝑛𝑚𝑎,𝑚0𝑎,𝑎,𝑎.
  • For any real constants 𝑛 and 𝑎, limprovidedandexist(𝑥+𝑎)𝑎𝑥=𝑛𝑎,𝑎𝑎.

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