Understanding the Algebra of Limits in Calculus

The algebra of limits formalises how the limit operation interacts with the basic algebraic operations of addition, subtraction, multiplication, and division for real-valued functions. These properties establish that the limit of a sum, difference, product, or quotient (under suitable conditions) is the sum, difference, product, or quotient of the limits. The precise theorems and their rigorous proofs are essential for solving limit problems systematically in calculus.

Statement of Algebraic Properties for Limits of Functions

Let $f(x)$ and $g(x)$ be functions defined on a punctured neighbourhood of $x = a$, and suppose that $\displaystyle \lim_{x \to a} f(x) = L$ and $\displaystyle \lim_{x \to a} g(x) = M$, where $L, M \in \mathbb{R}$. Then the following basic algebraic limit properties hold:

Sum Law: $\displaystyle \lim_{x \to a} [f(x) + g(x)] = L + M$

Difference Law: $\displaystyle \lim_{x \to a} [f(x) - g(x)] = L - M$

Constant Multiple Law: For any real constant $k$, $\displaystyle \lim_{x \to a} k f(x) = k L$

Product Law: $\displaystyle \lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$

Quotient Law: If $M \neq 0$, then $\displaystyle \lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{L}{M}$

Formal Proof of the Product Law for Limits of Functions

Let $\displaystyle \lim_{x \to a} f(x) = L$ and $\displaystyle \lim_{x \to a} g(x) = M$. To prove: $\displaystyle \lim_{x \to a} [f(x)g(x)] = LM$ using the $\epsilon\text{-}\delta$ definition of limits.

Step 1: For an arbitrary $\epsilon > 0$, we must find $\delta > 0$ such that

$0 < |x - a| < \delta \implies |f(x) g(x) - LM| < \epsilon$

Step 2: Expand $|f(x)g(x) - LM|$ as follows:

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

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

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

Step 3: Since $\displaystyle \lim_{x \to a} f(x) = L$, there exists $\delta_1 > 0$ such that if $0 < |x - a| < \delta_1$, then $|f(x) - L| < 1$.

Therefore, for such $x$, $|f(x)| \leq |L| + 1$. Denote $K = |L| + 1$ (so $f(x)$ is bounded near $a$).

Step 4: Let $\epsilon > 0$ be given. Choose $\eta_1 = \dfrac{\epsilon}{2(|M| + 1)}$ and $\eta_2 = \dfrac{\epsilon}{2K}$.

Since $\displaystyle \lim_{x \to a} f(x) = L$, there exists $\delta_2 > 0$ such that $0 < |x - a| < \delta_2 \implies |f(x) - L| < \eta_1$.

Since $\displaystyle \lim_{x \to a} g(x) = M$, there exists $\delta_3 > 0$ such that $0 < |x - a| < \delta_3 \implies |g(x) - M| < \eta_2$.

Step 5: Let $\delta = \min\{\delta_1, \delta_2, \delta_3\}$.

For all $x$ with $0 < |x - a| < \delta$, the following hold:

$|f(x)| \leq K$

$|f(x) - L| < \eta_1 = \dfrac{\epsilon}{2(|M| + 1)}$

$|g(x) - M| < \eta_2 = \dfrac{\epsilon}{2K}$

Step 6: Now,

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

$\leq K \cdot \dfrac{\epsilon}{2K} + |M| \cdot \dfrac{\epsilon}{2(|M| + 1)}$

$= \dfrac{\epsilon}{2} + \dfrac{|M|}{2(|M| + 1)}\epsilon$

$\leq \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2}$ (since $|M|/(|M| + 1) < 1$ for all $M \neq 0$ or $M = 0$)

$= \epsilon$

Conclusion: For all $x$ with $0 < |x-a| < \delta$, $|f(x)g(x) - LM| < \epsilon$. Thus, by the $\epsilon$–$\delta$ definition, $\lim_{x \to a} f(x)g(x) = LM$.

The other algebraic limit properties can be established similarly by explicit application of the $\epsilon$–$\delta$ definition. For a comprehensive treatment of limits, refer to the Limit Of A Function page.

Standard Results and Special Cases for Limits

For any constant $c \in \mathbb{R}$, $\displaystyle \lim_{x \to a} c = c$. For the identity function, $\displaystyle \lim_{x \to a} x = a$.

If $n$ is a positive integer, then $\displaystyle \lim_{x \to a} x^n = a^n$.

For any polynomial function $P(x)$, $\displaystyle \lim_{x \to a} P(x) = P(a)$. For a rational function $R(x) = \dfrac{P(x)}{Q(x)}$, if $Q(a) \neq 0$, then $\displaystyle \lim_{x \to a} R(x) = \dfrac{P(a)}{Q(a)}$.

Rigorous Evaluation of Limit Examples Using Algebraic Properties

Example 1: Compute $\displaystyle \lim_{x \to 2} (x^3 + 2x^2 + 4x - 2)$.

Given: $x^3 + 2x^2 + 4x - 2$

Substitution: Substitute $x = 2$ into each term.

$\lim_{x \to 2} (x^3 + 2x^2 + 4x - 2) = 2^3 + 2 \cdot 2^2 + 4 \cdot 2 - 2$

Simplification: $= 8 + 8 + 8 - 2$

$= 24 - 2$

$= 22$

Final Result: $\displaystyle \lim_{x \to 2} (x^3 + 2x^2 + 4x - 2) = 22$.

Example 2: Evaluate $\displaystyle \lim_{x \to 1} \dfrac{x^2 + 1}{x + 100}$.

Given: $\dfrac{x^2 + 1}{x + 100}$

Substitution: Substitute $x = 1$.

$\lim_{x \to 1} \dfrac{x^2 + 1}{x + 100} = \dfrac{1^2 + 1}{1 + 100}$

$= \dfrac{2}{101}$

Final Result: $\displaystyle \lim_{x \to 1} \dfrac{x^2 + 1}{x + 100} = \dfrac{2}{101}$.

For additional challenging examples, consult the Limits Solved Examples collection.

Application of Algebraic Limit Laws to Functions at Infinity

The algebraic properties remain valid for limits of functions as $x \to \infty$ (provided both individual limits exist). For rational functions $R(x) = \dfrac{P(x)}{Q(x)}$, consider the degrees $n = \deg P$, $m = \deg Q$:

If $n < m$, then $\displaystyle \lim_{x \to \infty} R(x) = 0$.

If $n = m$, then $\displaystyle \lim_{x \to \infty} R(x) = \dfrac{\text{leading coefficient of }P(x)}{\text{leading coefficient of }Q(x)}$.

If $n > m$, then the limit diverges (i.e., $+\infty$ or $-\infty$ as appropriate).

Formal Algebraic Properties for Limits of Sequences

If $(a_n)$ and $(b_n)$ are sequences such that $\lim_{n \to \infty} a_n = L$ and $\lim_{n \to \infty} b_n = M$, then:

Sum Law (Sequences): $\lim_{n \to \infty} (a_n + b_n) = L + M$

Product Law (Sequences): $\lim_{n \to \infty} (a_n b_n) = L M$

Quotient Law (Sequences): If $M \neq 0$, $\lim_{n \to \infty} \dfrac{a_n}{b_n} = \dfrac{L}{M}$

Explicit proofs for sequence limits are analogous to those for functions, replacing $\epsilon$–$\delta$ arguments with $\epsilon$–$N$ reasoning.

Summary of Key Algebra of Limits Results

The algebra of limits guarantees that fundamental operations such as addition, subtraction, multiplication, and division (when the denominator's limit is not zero) can be performed directly on limits of functions and sequences. This simplifies the evaluation of limits for common algebraic expressions, including polynomials and rational functions. For composite limit questions and rigorous continuity considerations, the algebraic limit theorems underpin all further developments in calculus. For deeper study, see Limit, Continuity And Differentiability.