5 Algebraic Techniques for Computing Limits

5.1 The Problem of Indeterminate Forms

The \varepsilon-\delta definition gives us a rigorous foundation for limits, but it offers little guidance for actually computing them. If we want to evaluate \lim_{x \to a} f(x), the definition tells us what it means for the limit to exist, but not how to find its value.

For many functions, the answer is straightforward. If f is a polynomial, we simply substitute: \lim_{x \to a} f(x) = f(a). The same holds for rational functions where the denominator doesn’t vanish, and for elementary functions like \sin x and e^x throughout their domains.

The difficulty arises when direct substitution yields an undefined expression. Consider

\lim_{x \to 1} \frac{x^2 - 1}{x - 1}, \quad \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}, \quad \lim_{x \to 0} \frac{\sin x}{x}.

Substituting x = 1 into the first expression gives \frac{0}{0}, which is undefined. Yet the limit exists—by factoring, we find it equals 2. These are indeterminate forms: the expression \frac{0}{0} is meaningless on its own, but the limit may still be well-defined.

This issue is fundamental to calculus. The derivative (one dimensional) itself can be defined as

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h},

which always has the form \frac{0}{0} when f is continuous at a. Understanding how to evaluate such limits is essential—without these techniques, we cannot compute derivatives.

In this chapter, we develop systematic methods for computing limits at finite points. We begin with the limit laws, which allow us to break complex expressions into simpler pieces. We then establish the squeeze theorem, which bounds difficult limits between known ones. Finally, we study algebraic techniques—factorization, rationalization, and trigonometric identities—that resolve indeterminate forms.

5.2 Limit Laws for Functions

In Section 3.2, we proved that limits of sequences respect algebraic operations: the limit of a sum is the sum of limits, the limit of a product is the product of limits, and so forth. These results hold because the limit operation preserves the algebraic structure of \mathbb{R}.

The same properties hold for function limits. The proofs are nearly identical to the sequence case, with the continuous parameter x replacing the discrete index n, and \delta replacing N.

Theorem 5.1 (Limit Laws for Functions) Let f, g : U \to \mathbb{R} with a a limit point of U. Suppose

\lim_{x \to a} f(x) = L, \qquad \lim_{x \to a} g(x) = M.

Then

\lim_{x \to a} [f(x) \pm g(x)] = L \pm M.
For any c \in \mathbb{R}, \lim_{x \to a} c f(x) = c L.
\lim_{x \to a} f(x) g(x) = L M.
If M \neq 0, then \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}.

Proof.

Let \varepsilon > 0.

By hypothesis, there exist \delta_1, \delta_2 > 0 such that

0 < |x - a| < \delta_1 \implies |f(x) - L| < \varepsilon/2,

0 < |x - a| < \delta_2 \implies |g(x) - M| < \varepsilon/2.

Set \delta = \min(\delta_1, \delta_2). Then for 0 < |x - a| < \delta,

|f(x) + g(x) - (L + M)| \le |f(x)-L| + |g(x)-M| < \varepsilon.

If c = 0, the result is trivial. If c \neq 0, choose \delta such that

0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon/|c|.

Then

|c f(x) - c L| = |c| \, |f(x)-L| < |c| \cdot \frac{\varepsilon}{|c|} = \varepsilon. Then

|c f(x) - c L| = |c| \, |f(x)-L| < |c| \cdot \frac{\varepsilon}{|c|} = \varepsilon.

We use the algebraic identity

f(x) g(x) - L M = f(x)[g(x) - M] + M[f(x) - L].

Taking absolute values and applying the triangle inequality,

|f(x) g(x) - L M| \le |f(x)| \, |g(x) - M| + |M| \, |f(x) - L|.

Since f(x) \to L, we can bound |f(x)|. Choose \delta_1 > 0 such that

0 < |x - a| < \delta_1 \implies |f(x) - L| < 1.

Then |f(x)| \le |f(x) - L| + |L| < 1 + |L| for such x.

Now choose \delta_2, \delta_3 > 0 such that

0 < |x - a| < \delta_2 \implies |g(x) - M| < \frac{\varepsilon}{2(1+|L|)},

0 < |x - a| < \delta_3 \implies |f(x) - L| < \frac{\varepsilon}{2(1+|M|)}.

Let \delta = \min(\delta_1, \delta_2, \delta_3). For 0 < |x - a| < \delta,

|f(x) g(x) - L M| \le (1+|L|) \cdot \frac{\varepsilon}{2(1+|L|)} + |M| \cdot \frac{\varepsilon}{2(1+|M|)} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.

It suffices to show \lim_{x \to a} \frac{1}{g(x)} = \frac{1}{M}, for then the product rule gives the full result.

Since M \neq 0, we can ensure g(x) stays bounded away from zero. Choose \delta_1 > 0 such that

0 < |x - a| < \delta_1 \implies |g(x) - M| < \frac{|M|}{2}.

By the triangle inequality, |g(x)| \ge |M| - |g(x) - M| > |M| - \frac{|M|}{2} = \frac{|M|}{2}, so g(x) doesn’t vanish near a.

Now observe

\left|\frac{1}{g(x)} - \frac{1}{M}\right| = \left|\frac{M - g(x)}{M g(x)}\right| = \frac{|g(x) - M|}{|M| \, |g(x)|} < \frac{|g(x) - M|}{|M| \cdot \frac{|M|}{2}} = \frac{2|g(x)-M|}{M^2}.

Choose \delta_2 > 0 such that

0 < |x - a| < \delta_2 \implies |g(x) - M| < \frac{\varepsilon M^2}{2}.

Let \delta = \min(\delta_1, \delta_2). Then for 0 < |x - a| < \delta,

\left|\frac{1}{g(x)} - \frac{1}{M}\right| < \frac{2 \cdot \frac{\varepsilon M^2}{2}}{M^2} = \varepsilon. \quad \square

5.3 Limits of Powers and Roots

Repeated application of Theorem 5.1 establishes limits for powers and roots.

Theorem 5.2 (Limits of Powers) If \lim_{x \to a} f(x) = L, then for any n \in \mathbb{N},

\lim_{x \to a} [f(x)]^n = L^n.

Proof.

Induction on n. For n = 1, the result is immediate.

Suppose the result holds for some n \ge 1. Then by the product rule (see Theorem 5.1, part 3),

\lim_{x \to a} [f(x)]^{n+1} = \lim_{x \to a} [f(x) \cdot f(x)^n] = \left(\lim_{x \to a} f(x)\right) \left(\lim_{x \to a} [f(x)]^n\right) = L \cdot L^n = L^{n+1}.

By induction, the result holds for all n \in \mathbb{N}. \square

Theorem 5.3 (Limits of Roots) If \lim_{x \to a} f(x) = L and n \in \mathbb{N}, then

\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L},

provided L > 0 when n is even, or L \ge 0 and f(x) \ge 0 in a neighborhood of a when n is odd.

Proof.

Let g(x) = \sqrt[n]{f(x)}. We must show \lim_{x \to a} g(x) = \sqrt[n]{L}.

Given \varepsilon > 0, we seek \delta > 0 such that |g(x) - \sqrt[n]{L}| < \varepsilon whenever 0 < |x - a| < \delta.

Since \lim_{x \to a} f(x) = L, there exists \delta_1 > 0 such that |f(x) - L| < \varepsilon' whenever 0 < |x - a| < \delta_1, for any \varepsilon' > 0 we specify.

For n even, assume L > 0. For n odd, assume L \ge 0 (the case L = 0 is handled separately).

Using the factorization f(x) - L = (\sqrt[n]{f(x)})^n - (\sqrt[n]{L})^n = (g(x) - \sqrt[n]{L}) \bigl([g(x)]^{n-1} + [g(x)]^{n-2}[\sqrt[n]{L}] + \cdots + [\sqrt[n]{L}]^{n-1}\bigr),

we have |f(x) - L| = |g(x) - \sqrt[n]{L}| \left| [g(x)]^{n-1} + [g(x)]^{n-2}[\sqrt[n]{L}] + \cdots + [\sqrt[n]{L}]^{n-1} \right|.

Choose \delta_2 small enough that |g(x)| < \sqrt[n]{L} + 1 for 0 < |x - a| < \delta_2. Then the sum is bounded by n(\sqrt[n]{L} + 1)^{n-1}. Thus |g(x) - \sqrt[n]{L}| \le \frac{|f(x) - L|}{n(\sqrt[n]{L} + 1)^{n-1}}.

Choosing \varepsilon' = \varepsilon \cdot n(\sqrt[n]{L} + 1)^{n-1} and \delta = \min\{\delta_1, \delta_2\} completes the argument. \square

Combining powers and roots yields rational exponents.

Theorem 5.4 (Limits of Rational Powers) If \lim_{x \to a} f(x) = L > 0 and r = m/n with m \in \mathbb{Z}, n \in \mathbb{N}, then

\lim_{x \to a} [f(x)]^r = L^r.

Proof.

Write [f(x)]^{m/n} = \sqrt[n]{[f(x)]^m}. By Theorem 5.2, \lim_{x \to a} [f(x)]^m = L^m. By Theorem 5.3, \lim_{x \to a} \sqrt[n]{[f(x)]^m} = \sqrt[n]{L^m} = L^{m/n}. \quad \square

Remark. Taking f(x) = x yields \lim_{x \to a} x^n = a^n for integer powers and \lim_{x \to a} x^r = a^r for rational powers (with a > 0 when necessary), which allows direct evaluation of polynomial and algebraic limits by substitution.

5.4 The Squeeze Theorem

When a limit cannot be evaluated directly, it often suffices to bound the function between two others whose limits are known and equal. This is the essence of the squeeze theorem.

Theorem 5.5 (Squeeze Theorem) Suppose f(x) \le g(x) \le h(x) in a deleted neighborhood of a. If

\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L,

then \lim_{x \to a} g(x) = L.

Proof.

Given \varepsilon > 0, choose \delta_1, \delta_2 > 0 such that

0 < |x - a| < \delta_1 \implies |f(x) - L| < \varepsilon, \quad 0 < |x - a| < \delta_2 \implies |h(x) - L| < \varepsilon.

Taking \delta = \min(\delta_1, \delta_2), we have for 0 < |x - a| < \delta:

L - \varepsilon < f(x) \le g(x) \le h(x) < L + \varepsilon.

Hence |g(x) - L| < \varepsilon. \square

Example 5.1 (Squeeze Theorem) Evaluate \displaystyle \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right).

Solution

The function \sin\left(\frac{1}{x}\right) oscillates wildly near x = 0, with no limit. However, since -1 \le \sin\left(\frac{1}{x}\right) \le 1, we have

-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2.

Both \pm x^2 \to 0 as x \to 0. By Theorem 5.5,

\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0. \quad \square

5.5 Factorization

If numerator and denominator both vanish at a, they admit (x-a) as a common factor.

Example 5.2 (Limit via Factoring) Evaluate \displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x - 1}.

Solution

For x \neq 1,

\frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x - 1} = x + 1.

Hence

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1) = 2. \quad \square

The following algebraic identities are frequently useful.

Lemma 5.1 (Difference of Powers) For n \in \mathbb{N},

x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + xa^{n-2} + a^{n-1}).

Proof.

Direct verification: expanding the product telescopes to x^n - a^n. \square

Lemma 5.2 (Sum of Odd Powers) For odd n,

x^n + a^n = (x + a)(x^{n-1} - x^{n-2}a + x^{n-3}a^2 - \cdots + (-1)^{n-2}xa^{n-2} + (-1)^{n-1}a^{n-1}).

Proof.

Set a \mapsto -a in Lemma 5.1. \square

Example 5.3 (Limit via Difference of Powers) Evaluate \displaystyle \lim_{x \to 2} \frac{x^5 - 32}{x - 2}.

Solution

By Lemma 5.1,

x^5 - 32 = x^5 - 2^5 = (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16).

Thus

\lim_{x \to 2} \frac{x^5 - 32}{x - 2} = \lim_{x \to 2} (x^4 + 2x^3 + 4x^2 + 8x + 16) = 16 + 16 + 16 + 16 + 16 = 80. \quad \square

5.6 Rationalization

Expressions involving radicals often succumb to multiplication by the conjugate, exploiting the identity (A - B)(A + B) = A^2 - B^2.

Definition 5.1 (Conjugate) The conjugate of A - B is A + B. Multiplication by \frac{A+B}{A+B} yields

(A - B) \cdot \frac{A + B}{A + B} = \frac{A^2 - B^2}{A + B}.

Example 5.4 (Limit via Conjugate) Evaluate \displaystyle \lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}.

Solution

Multiply by the conjugate:

\frac{\sqrt{1+x} - 1}{x} \cdot \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} = \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} = \frac{1}{\sqrt{1+x} + 1}.

Hence

\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} = \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1} = \frac{1}{2}. \quad \square

5.7 A Fundamental Trigonometric Limit

The limit

\lim_{x \to 0} \frac{\sin x}{x} = 1

is essential for differentiating trigonometric functions. The proof is geometric, comparing areas in the unit circle.

Theorem 5.6 (Trigonometric Limit) \lim_{x \to 0} \frac{\sin x}{x} = 1.

Proof

We establish the key inequality

\frac{1}{2} \sin \theta < \frac{1}{2} \theta < \frac{1}{2} \tan \theta.

in Theorem 3.5. Dividing by \frac{1}{2} \sin \theta > 0 gives

1 < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta}.

Taking reciprocals (reversing inequalities),

\cos \theta < \frac{\sin \theta}{\theta} < 1.

This inequality holds for \theta \in (0, \pi/2). Since \frac{\sin(-\theta)}{-\theta} = \frac{\sin \theta}{\theta}, it extends to \theta \in (-\pi/2, 0).

As \theta \to 0, we have \cos \theta \to 1. By Theorem 5.5,

\lim_{x \to 0} \frac{\sin x}{x} = 1. \quad \square

(For a detailed geometric construction with diagram, see the proof of the analogous sequence limit in Section 3.2.)

Corollary 5.1 \lim_{x \to 0} \frac{1 - \cos x}{x} = 0.

Proof

We use the conjugate to rationalize the numerator:

\frac{1 - \cos x}{x} = \frac{(1 - \cos x)(1 + \cos x)}{x(1 + \cos x)} = \frac{1 - \cos^2 x}{x(1 + \cos x)} = \frac{\sin^2 x}{x(1 + \cos x)}.

Rewrite this as

\frac{\sin^2 x}{x(1 + \cos x)} = \frac{\sin x}{x} \cdot \frac{\sin x}{1 + \cos x}.

As x \to 0, we have \frac{\sin x}{x} \to 1 by Theorem 5.6, and \frac{\sin x}{1 + \cos x} \to \frac{0}{2} = 0. Therefore

\lim_{x \to 0} \frac{1 - \cos x}{x} = 1 \cdot 0 = 0. \quad \square

Example 5.5 (Trigonometric Limit Variant) Evaluate \displaystyle \lim_{x \to 0} \frac{\sin 5x}{3x}.

Solution

Rewrite using Theorem 5.6

\frac{\sin 5x}{3x} = \frac{5}{3} \cdot \frac{\sin 5x}{5x}.

Let u = 5x. As x \to 0, we have u \to 0. Thus

\lim_{x \to 0} \frac{\sin 5x}{3x} = \frac{5}{3} \lim_{u \to 0} \frac{\sin u}{u} = \frac{5}{3} \cdot 1 = \frac{5}{3}. \quad \square

5.8 Summary of Algebraic Techniques

We conclude with an overview of techniques for computing limits at finite points.

Indeterminate Form	Technique	Example
\frac{0}{0}	Factor and cancel	\lim_{x \to 1} \frac{x^2-1}{x-1} = 2
\frac{0}{0} (roots)	Conjugate multiplication	\lim_{x \to 0} \frac{\sqrt{1+x}-1}{x} = \frac{1}{2}
Bounded oscillation	Squeeze theorem	\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0
Trigonometric	Use \lim_{x \to 0} \frac{\sin x}{x} = 1	\lim_{x \to 0} \frac{\sin 3x}{x} = 3