Lesson Explainer: Evaluating Limits Using Algebraic Techniques Mathematics • Higher Education

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

Specifically, we will cover the case where in seeking the limit limο—β†’οŒΊπ‘“(π‘₯),

  • the function is a rational function: 𝑓(π‘₯)=𝑃(π‘₯)𝑄(π‘₯), with 𝑃(π‘₯) and 𝑄(π‘₯) polynomials in π‘₯,
  • substitution fails because the quotient 𝑃(π‘Ž)𝑄(π‘Ž) gives the β€œindeterminate form” 00.

An example is lim→οŠͺπ‘₯βˆ’2π‘₯βˆ’56π‘₯βˆ’3π‘₯βˆ’4, where 𝑃(π‘₯)=π‘₯βˆ’2π‘₯βˆ’56 and 𝑄(π‘₯)=π‘₯βˆ’3π‘₯βˆ’4 give 𝑃(4)=64βˆ’(2)(4)βˆ’56=0 and 𝑄(4)=16βˆ’(3)(4)βˆ’4=0.

Since both 𝑃 and 𝑄 are polynomials, 𝑃(4)=𝑄(4)=0 means that (π‘₯βˆ’4) is a factor of each polynomial. So we can rewrite and divide these factors out: 𝑃(π‘₯)𝑄(π‘₯)=π‘₯βˆ’2π‘₯βˆ’56π‘₯βˆ’3π‘₯βˆ’4=(π‘₯βˆ’4)(π‘₯+4π‘₯+14)(π‘₯βˆ’4)(π‘₯+1)=π‘₯+4π‘₯+14π‘₯+1,π‘₯β‰ 4.providedthat

Now with the new function 𝑔(π‘₯)=π‘₯+4π‘₯+14π‘₯+1, we see that

  1. 𝑓(π‘₯)=𝑔(π‘₯) on the set of all numbers where both functions are defined, which is the set β„βˆ’{βˆ’1,4},
  2. the function 𝑔 is defined at π‘₯=4 (unlike 𝑓).

From (2), we can calculate the limit of 𝑔 at π‘₯=4 by substitution: lim→οŠͺοŠ¨π‘”(π‘₯)=𝑔(4)=(4)+4(4)+144+1=465.

Next, we use the following general principle.

Limits and Almost Equal Functions

Suppose that

  • functions 𝑓 and 𝑔 are equal at all points of an interval except π‘₯=π‘Ž,
  • 𝑔 has the limit 𝐿 as π‘₯ tends to π‘Ž.

Then, limο—β†’οŒΊπ‘“(π‘₯)=𝐿 also.

Now, fact (1) allows us to conclude that limlim→οŠͺ→οŠͺοŠ©οŠ¨π‘“(π‘₯)=π‘₯βˆ’2π‘₯βˆ’56π‘₯βˆ’3π‘₯βˆ’4=465.

Example 1: Finding the Limit of Rational Functions by Eliminating Common Factors

Find limο—β†’οŠ¨οŠ¨οŠ¨π‘₯βˆ’2π‘₯2π‘₯βˆ’6π‘₯+4.

Answer

Since (2)βˆ’2(2)=0 and 2(2)βˆ’6(2)+4=0 also, we need to factor out (π‘₯βˆ’2)π‘₯βˆ’2π‘₯2π‘₯βˆ’6π‘₯+4=(π‘₯βˆ’2)π‘₯(π‘₯βˆ’2)(2π‘₯βˆ’2)=π‘₯2π‘₯βˆ’2.

So limlimwhichcanbefoundbysubstitutionο—β†’οŠ¨οŠ¨οŠ¨ο—β†’οŠ¨π‘₯βˆ’2π‘₯2π‘₯βˆ’6π‘₯+4=π‘₯2π‘₯βˆ’2=22(2)βˆ’2=22=1.

The following factorization extends the difference of squares ο€Ήπ‘₯βˆ’π‘Žο…=(π‘₯βˆ’π‘Ž)(π‘₯+π‘Ž): π‘₯βˆ’π‘Ž=(π‘₯βˆ’π‘Ž)ο€Ήπ‘₯+π‘Žπ‘₯+π‘Žο…οŠ©οŠ©οŠ¨οŠ¨ and then ο€Ήπ‘₯βˆ’π‘Žο…=(π‘₯βˆ’π‘Ž)ο€Ήπ‘₯+π‘Žπ‘₯+π‘Žπ‘₯+π‘Žο…οŠͺοŠͺ and, more generally, π‘₯βˆ’π‘Ž=(π‘₯βˆ’π‘Ž)ο€Ήπ‘₯+π‘Žπ‘₯+π‘Žπ‘₯+β‹―+π‘Žπ‘₯+π‘Žο…οŠοŠοŠοŠ±οŠ§οŠοŠ±οŠ¨οŠ¨οŠοŠ±οŠ©οŠοŠ±οŠ¨οŠοŠ±οŠ§ which can be written as π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘₯+π‘Žπ‘₯+π‘Žπ‘₯+β‹―+π‘Žπ‘₯+π‘Ž, from which we deduce that limο—β†’οŒΊοŠοŠοŠοŠ±οŠ§π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘Ž since the right-hand side has 𝑛 terms.

This shortens potentially lengthy factorizations and has the following useful corollary: limο—β†’οŒΊοŠοŠο‰ο‰οŠοŠ±ο‰π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘šπ‘Ž.

Example 2: Finding the Limit of Rational Functions by Eliminating Common Factors

Find limο—β†’οŠ¨οŠ©οŠ¨8π‘₯βˆ’64π‘₯βˆ’4.

Answer

Substitution gives 00 and we note the common factor (π‘₯βˆ’2). Factoring the 8 in the numerator, 8π‘₯βˆ’64π‘₯βˆ’4=8ο€Ήπ‘₯βˆ’8π‘₯βˆ’4, reveals powers of 2: 8=2 and 4=2 so this is =8ο€Ήπ‘₯βˆ’2π‘₯βˆ’2.

Now limlimlimο—β†’οŠ¨οŠ©οŠ¨ο—β†’οŠ¨οŠ©οŠ©οŠ¨οŠ¨ο—β†’οŠ¨οŠ©οŠ©οŠ¨οŠ¨οŠ©οŠ±οŠ¨8π‘₯βˆ’64π‘₯βˆ’4=8ο€Ύπ‘₯βˆ’2π‘₯βˆ’2=8π‘₯βˆ’2π‘₯βˆ’2=8ο€Ό322=24, using the fact that limο—β†’οŒΊοŠοŠο‰ο‰οŠοŠ±ο‰π‘₯βˆ’π‘Žπ‘₯βˆ’π‘Ž=π‘›π‘šπ‘Ž.

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