Algebra of Limits

Introduction

1.The definition of a limit describes how a function $f(x)$ approaches a value near a point $x_0$, but on its own it is not a practical computational tool. In most problems we meet limits of functions that are added, multiplied, divided, composed, or raised to a power. The algebra of limits collects the rules that let a limit pass through each of these operations, every one of them derived from the definition. Throughout this page $L$ and $M$ denote real numbers with:

$$\lim_{x \to x_0} f(x) = L \qquad \lim_{x \to x_0} g(x) = M$$

Limit of a sum

2.When two functions approach finite values near a point, their sum approaches the sum of those values. If $f(x)$ stays close to $L$ and $g(x)$ stays close to $M$, their sum stays close to $L + M$:

$$\lim_{x \to x_0} \big( f(x) + g(x) \big) = L + M$$

3.The proof comes from the definition: given any tolerance around $L + M$, the deviations of $f(x)$ and $g(x)$ can be kept small enough, and separately, that their sum lands within the bound. The rule turns an awkward sum into two limits that are already known. Take

$$ \begin{align} f(x) &= \frac{\sin x}{x} \\[6pt] g(x) &= \frac{1 - \cos x}{x^2} \end{align} $$

4.neither of which is defined at the origin, together with the sum

$$\lim_{x \to 0} \big( f(x) + g(x) \big)$$

5.which substitution cannot reach. Each term is a standard remarkable limit:

$$ \begin{align} \lim_{x \to 0} \frac{\sin x}{x} &= 1 \\[6pt] \lim_{x \to 0} \frac{1 - \cos x}{x^2} &= \frac{1}{2} \end{align} $$

6.The sum rule then gives:

$$\lim_{x \to 0} \big( f(x) + g(x) \big) = 1 + \frac{1}{2} = \frac{3}{2}$$

Limit of a difference

7.The same argument applies to subtraction: if the two functions approach $L$ and $M$, their difference approaches $L - M$:

$$\lim_{x \to x_0} \big( f(x) - g(x) \big) = L - M$$

8.The proof repeats that of the sum, since $f - g$ is the sum of $f$ and $-g$. A more delicate case uses two quotients that both tend to $\frac{1}{2},$ so their difference is not settled by the two values on their own:

$$ \begin{align} f(x) &= \frac{1 - \cos x}{x^2} \\[6pt] g(x) &= \frac{\sin^2 x}{2x^2} \end{align} $$

9.Term by term the difference

$$\lim_{x \to 0} \big( f(x) - g(x) \big)$$

10.is undefined at $x = 0$, but both limits are known:

$$ \begin{align} \lim_{x \to 0} \frac{1 - \cos x}{x^2} &= \frac{1}{2} \\[6pt] \lim_{x \to 0} \frac{\sin^2 x}{2x^2} &= \frac{1}{2} \end{align} $$

11.The second follows from the remarkable limit $\lim_{x \to 0}\frac{\sin x}{x}=1.$ The difference rule gives:

$$\lim_{x \to 0} \big( f(x) - g(x) \big) = \frac{1}{2} - \frac{1}{2} = 0$$

12.Written as a single fraction

$$\frac{1 - \cos x}{x^2} - \frac{\sin^2 x}{2 x^2}$$

13.the value $0$ is far from obvious, since both terms tend to $\frac{1}{2}$ and only their cancellation produces it. Splitting the difference into two known limits sidesteps that cancellation.

Limit of a constant multiple

14.If a function approaches $L$, scaling it by a constant scales the limit by the same constant. This is the linearity of the limit in its simplest form: for any real constant $c$,

$$\lim_{x \to x_0} c f(x) = c L$$

15.The constant plays no part in the limiting process and only rescales the final value. A direct case is

$$\lim_{x \to 0} 3 \frac{\ln(1 + x)}{x}$$

16.The logarithmic quotient is undefined at $x = 0$, but the remarkable limit

$$\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1$$

17.supplies the limit of the unscaled quotient. Multiplying by the constant,

$$\lim_{x \to 0} 3 \cdot \frac{\ln(1 + x)}{x} = 3 \cdot 1 = 3$$

Limit of a product

18.If two functions approach finite values near a point, their product approaches the product of those values. With $f(x)$ close to $L$ and $g(x)$ close to $M$, the product stays close to $L \cdot M$:

$$\lim_{x \to x_0} \big( f(x) \cdot g(x) \big) = L \cdot M$$

19.Both factors stay bounded and can be controlled independently near $x_0$, so their product stays close to $L \cdot M$. Set

$$ \begin{align} f(x) &= \frac{e^x - 1}{x} \\[6pt] g(x) &= \frac{\ln(1 + x)}{x} \end{align} $$

20.Both are undefined at the origin, and both are remarkable limits:

$$ \begin{align} \lim_{x \to 0} \frac{e^x - 1}{x} &= 1 \\[6pt] \lim_{x \to 0} \frac{\ln(1 + x)}{x} &= 1 \end{align} $$

21.By the product rule:

$$\lim_{x \to 0} \frac{(e^x - 1) \ln(1 + x)}{x^2} = 1 \cdot 1 = 1$$

Limit of a quotient

22.When dividing two limits, a restriction on the denominator is unavoidable. If $g(x)$ approaches a nonzero value $M \neq 0$, the quotient behaves regularly near $x_0$. If the denominator tends to zero, the expression may be an indeterminate form of type $0/0$ or $\ell/0$ and needs separate analysis. Assuming $M \neq 0$:

$$\lim_{x \to x_0} \frac{f(x)}{g(x)} = \frac{L}{M}$$

23.Because $g(x)$ stays close to the nonzero number $M$, it is bounded away from zero on some neighbourhood of $x_0$, which excludes division by zero and lets the quotient behave as expected. Take

$$ \begin{align} f(x) &= \frac{e^x - 1}{x} \\[6pt] g(x) &= \frac{x^2 + 1}{x + 1} \end{align} $$

24.and their quotient as $x \to 0$:

$$\lim_{x \to 0} \frac{f(x)}{g(x)}$$

25.The numerator is undefined at $x = 0$ and comes from the remarkable limit:

$$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$

26.The denominator, continuous at $x = 0$, is handled by substitution:

$$\lim_{x \to 0} \frac{x^2 + 1}{x + 1} = \frac{0 + 1}{0 + 1} = 1$$

27.Since the denominator's limit is $M = 1 \neq 0$, the quotient rule applies and gives:

$$\lim_{x \to 0} \frac{f(x)}{g(x)} = \frac{1}{1} = 1$$

Limits of powers and polynomials

28.Taking $f = g$ in the product rule and iterating gives the limit of a power. If $f(x)$ approaches $L$, then for any positive integer $n$:

$$\lim_{x \to x_0} \big( f(x) \big)^n = L^n$$

29.A polynomial, built from sums and products, inherits these rules, so its limit is found by substituting the limiting value. For a power of a familiar quotient,

$$\lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)^4$$

30.the base is undefined at the origin but is again a remarkable limit:

$$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$

31.The power rule with $n = 4$ then gives:

$$\lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)^4 = 1^4 = 1$$

32.without expanding the fourth power, which would multiply the work for no gain.

Limit of a composition

33.The last rule concerns composition. For two functions $\varphi$ and $f$, the composition $\varphi(f(x))$ applies $f$ first and then $\varphi$ to the result. Suppose that:

$$\lim_{x \to x_0} f(x) = L$$

34.If $\varphi$ is continuous at $L$, the limit passes through the outer function:

$$\lim_{x \to x_0} \varphi \big( f(x) \big) = \varphi(L)$$

35.Continuity of $\varphi$ is what makes this work, since it forces small variations of the input near $L$ to produce small variations of the output. Without it the rule fails.

36.The square root of a remarkable limit is a simple instance. Set

$$f(x) = \frac{\ln(1 + x)}{x} \qquad \varphi(t) = \sqrt{t}$$

37.and compute:

$$\lim_{x \to 0} \sqrt{\frac{\ln(1 + x)}{x}}$$

38.The inner function is undefined at $x = 0$, with the remarkable limit:

$$\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1$$

39.Since $\varphi(t) = \sqrt{t}$ is continuous at $t = 1$, the limit passes through the square root:

$$\lim_{x \to 0} \sqrt{\frac{\ln(1 + x)}{x}} = \sqrt{1} = 1$$

Summary

40.The table below records the rules established above. Each holds when the individual limits $L$ and $M$ exist and the stated side conditions are met, in particular $M \neq 0$ for the quotient and continuity of $\varphi$ at $L$ for the composition.

42.When one or more of the underlying limits is infinite, or when the denominator of a quotient tends to zero, these rules do not apply directly.

43.The resulting expressions become indeterminate forms and require dedicated techniques such as factorisation, asymptotic comparison, L'Hôpital's rule, or Taylor expansions combined with little-o notation.