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Lesson 3 of 8

Limits and Derivatives · Lesson 3 of 8

Evaluating Algebraic Limits

“Use limit laws and justified simplification to calculate polynomial and rational limits.”

Learning Objectives

• Apply sum, difference, product, quotient, and constant-multiple laws with their conditions. • Explain why a polynomial limit is its value at the target. • Recognise that 0/0 is a signal to simplify rather than an answer. • Factor and combine rational expressions without changing nearby values. • Derive and apply the standard limit for a difference of powers.

From patterns to dependable laws

The previous lesson used tables and graphs to recognise what a function approaches. We now want a way to calculate the limit of an expression built from simpler functions. If the component limits exist as finite real numbers, the following laws let us combine them.

Suppose f(x) approaches L and g(x) approaches M as x approaches a. Then a sum approaches L+M, a difference approaches L−M, and a product approaches LM. A quotient approaches L/M only when M is non-zero. That last condition is a condition on the limiting denominator, not merely on the denominator at the sampled inputs.

Algebra of limitsLaTeX
Here L and M denote the finite limits of f and g at a. These laws are used without a formal proof in this chapter.

A fixed real number λ is a constant function with limit λ. The product law therefore gives a constant-multiple rule: multiplying a function by λ multiplies its limit by λ. This works for positive, negative, and zero constants.

Constant-multiple lawLaTeX
Common mistake

A quotient law cannot be used with limiting denominator zero. Writing 0/0 does not compute a limit: different expressions of that form can approach different values, or fail to have a finite limit.

Why substitution works for polynomials

A polynomial is a finite sum of terms consisting of constants times non-negative integer powers of x. We begin with the simple limit x→a. The product law gives x²→a², then x³→a³, and so on. Repeating this argument proves the corresponding result for every positive integer power.

Apply the constant-multiple rule to each term and the sum rule to the finitely many terms. Nothing has a denominator that could approach zero. The resulting number is exactly what direct substitution into the polynomial would give.

Definition
Polynomial function

A function of the form p(x)=a₀+a₁x+⋯+aₙxⁿ, where the coefficients are real constants and the highest non-zero coefficient determines the degree. Constant functions are included.

Polynomial limitLaTeX
Example — Three polynomial limits

Problem
Evaluate the limits of x³-x²+1 at 1, x(x+1) at 3, and 1+x+x²+⋯+x¹⁰ at −1.

  1. 1.For the first polynomial substitute x=1: 1³−1²+1=1.
  2. 2.For the second substitute x=3: 3(3+1)=12. Equivalently, the product law gives 3×4.
  3. 3.For the last expression, the signs alternate: 1−1+1−1+⋯+1. The five cancelling pairs leave one final 1.
  4. 4.The limits are respectively 1, 12, and 1. In each case the laws justify substitution.

Rational functions and a zero denominator

A rational function is a quotient of polynomials, defined only at inputs where its denominator is non-zero. If the denominator is also non-zero at the target a, the polynomial and quotient laws immediately justify substitution. If it becomes zero at a, inspect the expression further.

A common factor can sometimes be cancelled for inputs close to a but different from a. This does not redefine the original function at a. It gives a simpler expression with exactly the same nearby values, which is all a limit needs.

Definition
Rational function

A function g(x)/h(x) where g and h are polynomials, with the restriction h(x)≠0.

Direct substitution for a rational limitLaTeX
Example — A non-zero denominator

Problem
Evaluate lim[x→1] (x²+1)/(x+100).

  1. 1.The denominator at 1 is 101, so the quotient rule for limits is available.
  2. 2.The numerator limit is 1²+1=2 and the denominator limit is 1+100=101.
  3. 3.The required limit is 2/101.
Example — More zero factors in the numerator

Problem
Evaluate lim[x→2] (x³-4x²+4x)/(x²-4).

  1. 1.Factor the numerator as x(x-2)² and the denominator as (x-2)(x+2). Direct substitution initially produces 0/0.
  2. 2.For x≠2, cancel one factor x−2. The nearby expression is x(x-2)/(x+2).
  3. 3.Its denominator approaches 4 and its numerator approaches zero. The limit is 0.
  4. 4.One zero factor remains in the numerator after cancellation, which explains the zero result.
Example — A zero factor remains below

Problem
Evaluate lim[x→2] (x²-4)/(x³-4x²+4x).

  1. 1.After factoring and cancelling one x−2 for x≠2, the expression becomes (x+2)/[x(x-2)].
  2. 2.The numerator is close to 4 and the factor x is close to 2. The remaining factor x−2 changes sign across 2 and approaches zero.
  3. 3.Thus the values become large negative numbers from the left and large positive numbers from the right. No finite two-sided limit exists.
  4. 4.A leftover denominator zero cannot be resolved by simply writing 4/0.
Example — Equal numbers of zero factors

Problem
Evaluate lim[x→2] (x³-2x²)/(x²-5x+6).

  1. 1.The numerator is x²(x-2) and the denominator is (x-2)(x-3).
  2. 2.For x≠2, the quotient equals x²/(x-3).
  3. 3.The new denominator tends to −1, so substitution gives 4/(−1)=−4.
Example — Combine before cancelling

Problem
Evaluate lim[x→1] [(x-2)/(x²-x)-1/(x³-3x²+2x)].

  1. 1.Factor the denominators as x(x-1) and x(x-1)(x-2). Use the common denominator x(x-1)(x-2).
  2. 2.The numerator becomes (x-2)²-1=x²-4x+3=(x-1)(x-3).
  3. 3.For nearby allowed inputs cancel x−1 to obtain (x-3)/[x(x-2)].
  4. 4.Substitute x=1 into this simplified expression: (1−3)/[1(1−2)]=2.
  5. 5.Each original term has a vanishing denominator at 1, so evaluating the two terms separately would not justify the difference. Combining them exposes the cancellation.

We can organise these cases by counting powers of x−a. Write g(x)=(x−a)ᵏg₁(x) and h(x)=(x−a)ˡh₁(x), where g₁(a) and h₁(a) are non-zero. After common factors are removed, the comparison of k and l tells us what remains. This is a way of explaining the examples, not an instruction to apply the quotient law to zero denominators.

ComparisonWhat remains near aFinite-limit conclusion
k>lA positive power of x−a in the numeratorLimit 0
k=lg₁(x)/h₁(x)Limit g₁(a)/h₁(a)
k<lA positive power of x−a in the denominatorNo finite real limit

A difference of powers gives a reusable limit

The expression (xⁿ−aⁿ)/(x−a) is especially important because its numerator and denominator both vanish at a. For a positive integer n, a finite factorisation removes the common factor. The simplified expression contains n terms, each approaching aⁿ⁻¹.

Difference-of-powers factorisationLaTeX

For x≠a, divide by x−a and take the limit term by term. The first term tends to aⁿ⁻¹. The second tends to a·aⁿ⁻²=aⁿ⁻¹, and each remaining term has the same limiting value. Adding n equal terms gives the result below. For n=1, the original quotient is simply 1.

Standard power limitLaTeX
The chapter also states the corresponding result for rational n when a>0; that extension is not proved here.
Example — A ratio of two differences of powers

Problem
Evaluate lim[x→1] (x¹⁵-1)/(x¹⁰-1).

  1. 1.For x≠1, divide both numerator and denominator by x−1. This gives a quotient of the two standard power-limit expressions.
  2. 2.The top expression has limit 15×1¹⁴=15. The bottom has limit 10×1⁹=10.
  3. 3.The limiting denominator is non-zero, so their quotient has limit 15/10=3/2.
Example — A square-root limit

Problem
Evaluate lim[x→0] (√(1+x)-1)/x.

  1. 1.Put y=1+x. Then y→1, and the expression is (y^(1/2)-1)/(y-1). The rational-power extension gives the limit 1/2.
  2. 2.To see the result directly, multiply numerator and denominator by √(1+x)+1. The numerator becomes (1+x)−1=x.
  3. 3.For non-zero x the expression becomes 1/(√(1+x)+1), which approaches 1/(1+1)=1/2.
  4. 4.This alternate derivation explains the cancellation without relying on the unproved extension.

Choose a method before calculating

First identify the function and target. If substitution produces a defined number, the relevant laws often finish the calculation. If it produces 0/0, look for factors, a common denominator, or a difference-of-powers structure. If a denominator zero remains with a non-zero numerator, do not force a finite answer.

Keep the original domain restrictions even after simplifying. In a limit calculation the missing target can be harmless; in a question asking for the function value it cannot be ignored. This difference is exactly what makes algebraic simplification useful for limits.

Quiz

Quick check

When is the quotient law directly valid?

Quick check

What is the limit of x³−x²+1 at x=1?

Quick check

Why may x−a be cancelled in a limit as x→a?

Quick check

What is the limit of (x⁴−1)/(x−1) at 1?

Quick check

What does the form 0/0 tell us?

Quick check

If cancellation leaves (x−a)² in the numerator and a non-zero limiting denominator, what is the limit?

Practice Problems

Practice Problems
  1. Evaluate lim[x→3] (x+3) and explain the law used.
  2. Evaluate lim[x→π] (x-22/7). Keep π exact.
  3. Evaluate lim[r→1] π r² and identify which quantity varies.
  4. Evaluate lim[x→4] (4x+3)/(x-2).
  5. Evaluate lim[x→-1] (x¹⁰+x⁵+1)/(x-1).
  6. Evaluate lim[x→0] [(x+1)⁵-1]/x by expanding or using the standard power limit after a substitution.
  7. Evaluate lim[x→2] (3x²-x-10)/(x²-4) by factorisation.
  8. Evaluate lim[x→3] (x⁴-81)/(2x²-5x-3).
  9. Evaluate lim[x→0] (ax+b)/(cx+1) for fixed real constants a, b, c.
  10. Evaluate lim[z→1] (z^(1/3)-1)/(z^(1/6)-1) and justify the change of variable or power-limit method.
  11. Evaluate lim[x→1] (ax²+bx+c)/(cx²+bx+a) when a+b+c is non-zero.
  12. Evaluate lim[x→-2] (1/x+1/2)/(x+2) by writing the numerator as one fraction.
  13. For f(x)=∏[i=1 to n](x-aᵢ) with fixed real constants, find the limit at a₁ and at a number a different from every aᵢ.
  14. If lim[x→1] [f(x)-2]/(x²-1)=π, determine lim[x→1] f(x). Rewrite f(x) using the given quotient before applying the product law.

Key Takeaways

Key Takeaways

• Limit laws apply to existing finite component limits, and a quotient needs a non-zero limiting denominator. • Polynomial limits equal the substituted polynomial value. • A 0/0 form requires simplification and is not itself an answer. • Cancelling common factors preserves nearby values while retaining the original domain restrictions. • The standard difference-of-powers limit follows from factorisation and n equal limiting terms.

Previous · Lesson 2

Understanding Limits from Both Sides

Next · Lesson 4

Trigonometric Limits and the Squeeze Idea