Mathematics · Ch 12 — Limits and Derivatives

Algebra of Limits

12.3.1

Algebra of Limits

The Algebra of Limits: Why Limits Behave So Nicely

When you worked through the earlier illustrations in this chapter, you probably noticed something convenient: the process of taking a limit seemed to "play well" with addition, subtraction, multiplication, and division. If you knew the limits of two separate functions, you could almost always predict the limit of their sum, difference, product, or quotient. This wasn't just luck — it's a fundamental property of limits that holds whenever the individual limits exist and are finite.

The key idea is that limits respect arithmetic. If two functions each approach a specific number as $x$ gets close to $a$ , then their sum approaches the sum of those numbers, their product approaches the product, and so on. This is what makes evaluating limits manageable: instead of wrestling with complicated expressions directly, you can break them into simpler pieces, find the limit of each piece, and then combine the results.

Important

The algebra of limits only works when both $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist (and, for division, when the denominator's limit is non-zero). If either limit does not exist, these rules do not apply.

Theorem 1: The Four Fundamental Limit Laws

The textbook formalises these observations into a single theorem. It states the result without proof — the proof is typically covered in more advanced calculus courses — but the statement itself is your working toolkit.

f

Limit of a sum The limit of the sum of two functions equals the sum of their individual limits. $\lim_{x \to a} \big[ f(x) + g(x) \big] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$
Limit of a difference The limit of the difference of two functions equals the difference of their individual limits. $\lim_{x \to a} \big[ f(x) - g(x) \big] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$
Limit of a product The limit of the product of two functions equals the product of their individual limits. $\lim_{x \to a} \big[ f(x) \cdot g(x) \big] = \left( \lim_{x \to a} f(x) \right) \cdot \left( \lim_{x \to a} g(x) \right)$
Limit of a quotient The limit of the quotient of two functions equals the quotient of their individual limits, provided the limit of the denominator is not zero. $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\displaystyle \lim_{x \to a} f(x)}{\displaystyle \lim_{x \to a} g(x)}, \quad \text{provided } \lim_{x \to a} g(x) \neq 0$
Watch out

Property (iv) is the trickiest. If $\lim_{x \to a} g(x) = 0$ , you cannot simply divide the limits. The expression $\frac{f(x)}{g(x)}$ may still have a limit, but you must handle it using other methods (like factoring or rationalisation) — the quotient rule simply does not apply here.

A Special Case: Multiplying by a Constant

There is an important special case of property (iii) that deserves its own spotlight. Suppose $g$ is a constant function — that is, $g(x) = \lambda$ for some fixed real number $\lambda$ , no matter what $x$ is. Then $\lim_{x \to a} g(x) = \lambda$ , and property (iii) becomes:

\lim_{x \to a} \big[ \lambda \cdot f(x) \big] = \lambda \cdot \lim_{x \to a} f(x)

This is extremely useful: you can always "pull a constant factor" out in front of a limit. It means that scaling a function by a constant simply scales its limit by the same constant.

Tip

This constant-multiple rule is so common that you will use it in almost every limit problem. For example, $\lim_{x \to 2} 5x^2 = 5 \cdot \lim_{x \to 2} x^2 = 5 \cdot 4 = 20$ .

How These Laws Are Used (A Preview) …

Theorem 1

Theorem 1: Algebra of Limits

Let $f$ and $g$ be two functions such that both $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist. Then:

Limit of a sum: $\displaystyle \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$
Limit of a difference: $\displaystyle \lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$
Limit of a product: $\displaystyle \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$
Limit of a quotient: $\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$ , provided $\lim_{x \to a} g(x) \neq 0$

Important

The theorem only applies when both individual limits exist. If either limit does not exist, these rules cannot be applied directly.

Note

A special case of part (iii): when $g$ is a constant function $g(x) = \lambda$ for some real number $\lambda$ , we get $\displaystyle \lim_{x \to a} [\lambda \cdot f(x)] = \lambda \cdot \lim_{x \to a} f(x)$ . This is often called the "constant multiple rule."

›Proof

Proof of Theorem 1

Since $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist, let us denote:

$\displaystyle \lim_{x \to a} f(x) = L$ and $\displaystyle \lim_{x \to a} g(x) = M$

Part (i): Limit of a sum

We need to show that $\displaystyle \lim_{x \to a} [f(x) + g(x)] = L + M$ .

By the definition of limit, for any $\epsilon > 0$ , there exists $\delta_1 > 0$ such that whenever $0 < |x - a| < \delta_1$ , we have $|f(x) - L| < \frac{\epsilon}{2}$ .

Similarly, there exists $\delta_2 > 0$ such that whenever $0 < |x - a| < \delta_2$ , we have $|g(x) - M| < \frac{\epsilon}{2}$ .

Choose $\delta = \min(\delta_1, \delta_2)$ . Then for $0 < |x - a| < \delta$ , both inequalities hold simultaneously. Now:

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

By the triangle inequality: $|(f(x) - L) + (g(x) - M)| \leq |f(x) - L| + |g(x) - M|$

$< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Thus, for any $\epsilon > 0$ , we have found a $\delta > 0$ such that $0 < |x - a| < \delta$ implies $|[f(x) + g(x)] - (L + M)| < \epsilon$ . This proves $\displaystyle \lim_{x \to a} [f(x) + g(x)] = L + M$ .

Part (ii): Limit of a difference

We need to show $\displaystyle \lim_{x \to a} [f(x) - g(x)] = L - M$ .

Observe that $[f(x) - g(x)] - (L - M) = (f(x) - L) - (g(x) - M)$ .

Using the same $\delta_1, \delta_2$ as in part (i) and choosing $\delta = \min(\delta_1, \delta_2)$ :

$|[f(x) - g(x)] - (L - M)| = |(f(x) - L) - (g(x) - M)|$

$\leq |f(x) - L| + |g(x) - M|$ (by the triangle inequality)

$< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Hence $\displaystyle \lim_{x \to a} [f(x) - g(x)] = L - M$ .

Part (iii): Limit of a product

We need to show $\displaystyle \lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$ .

Consider:

$|f(x)g(x) - LM| = |f(x)g(x) - f(x)M + f(x)M - LM|$

$= |f(x)(g(x) - M) + M(f(x) - L)|$

$\leq |f(x)||g(x) - M| + |M||f(x) - L|$

Since $\lim_{x \to a} f(x) = L$ , there exists $\delta_1 > 0$ such that for $0 < |x - a| < \delta_1$ , $|f(x) - L| < 1$ , which implies $|f(x)| < |L| + 1$ .

Also, for any $\epsilon > 0$ , there exists $\delta_2 > 0$ such that for $0 < |x - a| < \delta_2$ , $|f(x) - L| < \frac{\epsilon}{2(|M| + 1)}$ .

And there exists $\delta_3 > 0$ such that for $0 < |x - a| < \delta_3$ , $|g(x) - M| < \frac{\epsilon}{2(|L| + 1)}$ .

Choose $\delta = \min(\delta_1, \delta_2, \delta_3)$ . Then for $0 < |x - a| < \delta$ :

$|f(x)g(x) - LM| \leq |f(x)||g(x) - M| + |M||f(x) - L|$

$< (|L| + 1) \cdot \frac{\epsilon}{2(|L| + 1)} + |M| \cdot \frac{\epsilon}{2(|M| + 1)}$

$< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Therefore $\displaystyle \lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$ .

Part (iv): Limit of a quotient

We need to show $\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$ , provided $M \neq 0$ .

First, we show $\displaystyle \lim_{x \to a} \frac{1}{g(x)} = \frac{1}{M}$ .

Since $M \neq 0$ , there exists $\delta_1 > 0$ such that for $0 < |x - a| < \delta_1$ , $|g(x) - M| < \frac{|M|}{2}$ , which implies $|g(x)| > \frac{|M|}{2}$ .

For any $\epsilon > 0$ , there exists $\delta_2 > 0$ such that for $0 < |x - a| < \delta_2$ , $|g(x) - M| < \frac{\epsilon |M|^2}{2}$ .

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

$\left|\frac{1}{g(x)} - \frac{1}{M}\right| = \left|\frac{M - g(x)}{g(x)M}\right| = \frac{|g(x) - M|}{|g(x)||M|}$

$< \frac{|g(x) - M|}{\frac{|M|}{2} \cdot |M|} = \frac{2|g(x) - M|}{|M|^2}$

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