Limit Laws and Algebraic Techniques

Learning objectives

By the end of this lesson, you should be able to:

1. use the sum, product, quotient, and composition laws for limits;
2. distinguish an indeterminate form from a nonexistent limit;
3. choose an algebraic transformation that preserves nearby behavior; and
4. justify each step of a limit calculation.

The central idea

In the previous lesson, we replaced the intuitive language of limits with the formal epsilon-delta definition. We can now use that definition to establish the algebraic laws that govern limits and then apply those laws to calculations.

A limit concerns the trend of a function near a point. The value of the function at the point may matter for other questions, but the limit is determined by what the function approaches as the input approaches the target.

This makes the algebra of limits fundamental. The sum, difference, product, reciprocal, and quotient laws are not merely computational rules: they are consequences of the formal definition.

Suppose

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

The sum and difference laws

For a sum,

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

Likewise,

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

The intermediate step is important: the limit of a sum is the sum of the individual limits. We are not merely substituting $L$ and $M$ into a formula; we are using a limit law that permits the original limit to be separated into simpler limits.

Preparing to prove the limit laws

The previous lesson established the formal definition of a limit. We can therefore do more than state the limit laws: we can prove them.

We will prove the product law in this lesson because its proof introduces an important additional idea: a function with a finite limit can be bounded in a sufficiently small neighborhood of the limiting point. After studying that proof, you will be asked in Test Your Knowledge to use the formal epsilon-delta definition to prove the sum law yourself.

The sum-law proof is simpler than the product-law proof. It does not require an auxiliary bound on either function, but it does require the triangle inequality. Recall that the triangle inequality states that for any real numbers $u$ and $v$,

$$|u+v|\le |u|+|v|.$$

In words, the absolute value of a sum cannot exceed the sum of the absolute values.

A constant multiple behaves similarly. For a real constant $c$,

$$\lim_{x\to a}[c f(x)] = c\lim_{x\to a}f(x) = cL.$$

These facts can be combined. For example,

$$\lim_{x\to a}[3f(x)-2g(x)] = 3\lim_{x\to a}f(x) - 2\lim_{x\to a}g(x) = 3L-2M.$$

The product law

For a product,

$$\lim_{x\to a}[f(x)g(x)] = \left(\lim_{x\to a}f(x)\right) \left(\lim_{x\to a}g(x)\right) = LM.$$

Proof of the product law

Suppose

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

We want to prove

$$\lim_{x\to a}[f(x)g(x)]=LM.$$

Let $\varepsilon>0$ be arbitrary. We need to control

$$|f(x)g(x)-LM|.$$

The expression does not yet separate naturally into the two quantities controlled by the hypotheses,

$$|f(x)-L| \qquad\text{and}\qquad |g(x)-M|.$$

To create those differences, insert the same intermediate term $Lg(x)$ by adding and subtracting it:

$$\begin{aligned} f(x)g(x)-LM &= f(x)g(x)-Lg(x)+Lg(x)-LM\\ &= g(x)[f(x)-L]+L[g(x)-M]. \end{aligned}$$

Taking absolute values and applying the triangle inequality gives

$$\begin{aligned} |f(x)g(x)-LM| &= |g(x)[f(x)-L]+L[g(x)-M]|\\ &\le |g(x)||f(x)-L| + |L||g(x)-M|. \end{aligned}$$

The second term can be controlled directly through $|g(x)-M|$. The first term contains the additional factor $|g(x)|$, so we must first establish a bound on $g(x)$ near $a$.

Because

$$\lim_{x\to a}g(x)=M,$$

the formal definition applies for every positive output tolerance. In particular, choose the convenient output tolerance $1$. Then there exists $\delta_0>0$ such that

$$0<|x-a|<\delta_0 \Longrightarrow |g(x)-M|<1.$$

Under this restriction,

$$\begin{aligned} |g(x)| &= |g(x)-M+M|\\ &\le |g(x)-M|+|M|\\ &< 1+|M|. \end{aligned}$$

Thus, the local bound

$$|g(x)|<1+|M|$$

has been established whenever $0<|x-a|<\delta_0$. It is not being asserted as a global bound on $g$.

We now control the two terms in the estimate. Since

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

there exists $\delta_1>0$ such that

$$0<|x-a|<\delta_1 \Longrightarrow |f(x)-L| < \frac{\varepsilon}{2(1+|M|)}.$$

Likewise, since

$$\lim_{x\to a}g(x)=M,$$

there exists $\delta_2>0$ such that

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

Choose

$$\delta=\min(\delta_0,\delta_1,\delta_2).$$

Now suppose

$$0<|x-a|<\delta.$$

Because $\delta\le\delta_0$, the local bound $|g(x)|<1+|M|$ applies. Because $\delta\le\delta_1$ and $\delta\le\delta_2$, the two epsilon-dependent bounds also apply. Therefore,

$$\begin{aligned} |f(x)g(x)-LM| &\le |g(x)||f(x)-L| + |L||g(x)-M|\\ &< (1+|M|) \frac{\varepsilon}{2(1+|M|)} + |L| \frac{\varepsilon}{2(1+|L|)}\\ &\le \frac{\varepsilon}{2} + \frac{\varepsilon}{2}\\ &= \varepsilon. \end{aligned}$$

Therefore,

$$0<|x-a|<\delta \Longrightarrow |f(x)g(x)-LM|<\varepsilon.$$

By the formal definition of a limit,

$$\boxed{ \lim_{x\to a}[f(x)g(x)] = \left(\lim_{x\to a}f(x)\right) \left(\lim_{x\to a}g(x)\right) = LM. }$$

The product law requires more care than the sum law because the product introduces a factor whose magnitude must itself be controlled near $a$. The local bound on $g(x)$ is what makes the epsilon-dependent estimate possible.

This law immediately gives us the power law for positive integers. Since

$$[f(x)]^n = \underbrace{f(x)f(x)\cdots f(x)}_{n\text{ factors}},$$

we may apply the product law repeatedly:

$$\begin{aligned} \lim_{x\to a}[f(x)]^n &= \lim_{x\to a} \left[ \underbrace{f(x)f(x)\cdots f(x)}_{n\text{ factors}} \right] \\ &= \underbrace{ \left(\lim_{x\to a}f(x)\right) \left(\lim_{x\to a}f(x)\right) \cdots \left(\lim_{x\to a}f(x)\right) }_{n\text{ factors}} \\ &= L^n. \end{aligned}$$

Thus the power law is not an isolated rule. It follows directly from repeated application of the product law.

The reciprocal and quotient laws

Division requires additional care. First consider the reciprocal of $g$. If

$$\lim_{x\to a}g(x)=M$$

and $M\neq0$, then

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

Now write a quotient as multiplication by a reciprocal:

$$\frac{f(x)}{g(x)} = f(x)\frac{1}{g(x)}.$$

Using the product law and then the reciprocal law,

$$\begin{aligned} \lim_{x\to a}\frac{f(x)}{g(x)} &= \lim_{x\to a} \left[ f(x)\frac{1}{g(x)} \right] \\ &= \left(\lim_{x\to a}f(x)\right) \left(\lim_{x\to a}\frac{1}{g(x)}\right) \\ &= L\left(\frac{1}{M}\right) \\ &= \frac{L}{M}, \qquad M\neq0. \end{aligned}$$

The condition $M\neq0$ is essential. If the denominator approaches zero, the quotient law does not apply. We must investigate the expression further rather than treating division by zero as a valid algebraic step.

Combining the laws

These laws are most useful when used together. If

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

with $N\neq0$, then

$$\begin{aligned} \lim_{x\to a}\frac{f(x)+g(x)}{h(x)} &= \frac{\displaystyle\lim_{x\to a}[f(x)+g(x)]} {\displaystyle\lim_{x\to a}h(x)} \\ &= \frac{\displaystyle\lim_{x\to a}f(x)+\lim_{x\to a}g(x)} {\displaystyle\lim_{x\to a}h(x)} \\ &= \frac{L+M}{N}. \end{aligned}$$

Each equality is justified by a specific limit law. Writing the intermediate steps makes that logic visible.

For roots and other continuous outer functions, the limiting value can also be passed through the outer function when the resulting value lies in the appropriate domain. We will use that fact below when working with radicals.

Indeterminate forms

Substitution may produce $\frac{0}{0}$. That symbol is an indeterminate form: it says the quotient law cannot yet be used because its denominator condition fails. It does not say the limit is zero, infinite, or nonexistent.

The notation $\frac{0}{0}$ is therefore not the value of the limit. It is a diagnostic result from substitution.

For example, as $x\to0$, each of

$$\frac{x}{x},\qquad \frac{x^2}{x},\qquad \frac{x}{x^2}$$

produces $\frac{0}{0}$ by direct substitution, yet their behaviors are very different:

$$\lim_{x\to0}\frac{x}{x}=1,$$ $$\lim_{x\to0}\frac{x^2}{x}=0.$$

For the third expression,

$$\lim_{x\to0}\frac{x}{x^2}$$

does not exist as a finite real limit.

The limit does not exist as a finite real limit because $\frac{1}{x}$ decreases without bound as $x\to0^-$ and increases without bound as $x\to0^+$. This is the same distinction between an indeterminate form and actual function behavior that we encountered in the previous lesson.

Three useful transformations

An indeterminate form tells us that direct substitution has not finished the problem. Algebra can often rewrite the function into an equivalent form for nearby values of $x$ so that the limit laws can then be applied.

Factoring and cancellation

Evaluate

$$\lim_{x\to a}\frac{x^2-a^2}{x-a}.$$

Direct substitution produces $\frac{0}{0}$. Factor the difference of squares:

$$\begin{aligned} \lim_{x\to a}\frac{x^2-a^2}{x-a} &= \lim_{x\to a}\frac{(x-a)(x+a)}{x-a}. \end{aligned}$$

For $x\neq a$, the common factor $x-a$ may be canceled:

$$\frac{(x-a)(x+a)}{x-a} = x+a.$$

Therefore,

$$\begin{aligned} \lim_{x\to a}\frac{x^2-a^2}{x-a} &= \lim_{x\to a}(x+a) \\ &= \lim_{x\to a}x+\lim_{x\to a}a \\ &= a+a \\ &= 2a. \end{aligned}$$

The cancellation does not assert that the original quotient is defined at $x=a$. It shows that the original quotient and $x+a$ have the same values for nearby inputs $x\neq a$, which is exactly the behavior relevant to the limit.

Rationalizing with a conjugate

Evaluate

$$\lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}.$$

Direct substitution produces $\frac{0}{0}$. Multiply by the conjugate of the numerator:

$$\begin{aligned} \lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4} &= \lim_{x\to4} \left[ \frac{\sqrt{x+5}-3}{x-4} \cdot \frac{\sqrt{x+5}+3}{\sqrt{x+5}+3} \right] \\ &= \lim_{x\to4} \frac{(\sqrt{x+5}-3)(\sqrt{x+5}+3)} {(x-4)(\sqrt{x+5}+3)}. \end{aligned}$$

Use the difference-of-squares identity:

$$(\sqrt{x+5}-3)(\sqrt{x+5}+3) = (x+5)-9 = x-4.$$

Thus, for $x\neq4$,

$$\frac{(\sqrt{x+5}-3)(\sqrt{x+5}+3)} {(x-4)(\sqrt{x+5}+3)} = \frac{x-4}{(x-4)(\sqrt{x+5}+3)} = \frac{1}{\sqrt{x+5}+3}.$$

Now the denominator approaches $6$, not $0$, so the quotient law applies:

$$\begin{aligned} \lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4} &= \lim_{x\to4}\frac{1}{\sqrt{x+5}+3} \\ &= \frac{1}{\displaystyle\lim_{x\to4}(\sqrt{x+5}+3)} \\ &= \frac{1}{\sqrt{9}+3} \\ &= \frac16. \end{aligned}$$

Combining fractions

Evaluate

$$\lim_{x\to a} \frac{\frac1x-\frac1a}{x-a}, \qquad a\neq0.$$

The numerator must first be combined into a single fraction. Use the common denominator $ax$:

$$\frac1x = \frac{a}{ax}, \qquad \frac1a = \frac{x}{ax}.$$

Therefore,

$$\frac1x-\frac1a = \frac{a}{ax}-\frac{x}{ax} = \frac{a-x}{ax}.$$

Substitute this result into the original expression:

$$\begin{aligned} \frac{\frac1x-\frac1a}{x-a} &= \frac{\frac{a-x}{ax}}{x-a} \\ &= \frac{a-x}{ax(x-a)}. \end{aligned}$$

Since

$$a-x=-(x-a),$$

we obtain

$$\begin{aligned} \frac{a-x}{ax(x-a)} &= \frac{-(x-a)}{ax(x-a)} \\ &= -\frac1{ax}, \qquad x\neq a. \end{aligned}$$

The limit is therefore

$$\begin{aligned} \lim_{x\to a} \frac{\frac1x-\frac1a}{x-a} &= \lim_{x\to a}\left(-\frac1{ax}\right) \\ &= -\frac{1}{a\displaystyle\lim_{x\to a}x} \\ &= -\frac1{a^2}, \qquad a\neq0. \end{aligned}$$

Worked example

Evaluate

$$\lim_{x\to2}\frac{x^3-8}{x^2-4}.$$

Direct substitution gives $\frac{0}{0}$, so factor both differences:

$$x^3-8=(x-2)(x^2+2x+4),\qquad x^2-4=(x-2)(x+2).$$

Substitute the factored forms into the quotient:

$$\frac{x^3-8}{x^2-4} = \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}.$$

For $x\neq2$, cancel the common factor $x-2$:

$$\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = \frac{x^2+2x+4}{x+2}.$$

The simplified denominator tends to $4\neq0$, so the quotient law now applies:

$$\lim_{x\to2}\frac{x^3-8}{x^2-4} =\frac{4+4+4}{4}=3.$$

Notice the logic: substitution diagnosed the obstacle; factoring removed a common factor; only then did the limit laws finish the calculation.

Common mistakes

Test Your Knowledge

1. Apply the limit laws using functional notation

Suppose

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

where $N\neq0$.

Evaluate

$$\lim_{x\to a}\frac{f(x)+g(x)}{h(x)}.$$

Show the intermediate limit-law steps.

Solution

Apply the quotient law first, then the sum law:

$$\begin{aligned} \lim_{x\to a}\frac{f(x)+g(x)}{h(x)} &= \frac{\displaystyle\lim_{x\to a}[f(x)+g(x)]} {\displaystyle\lim_{x\to a}h(x)} \\ &= \frac{\displaystyle\lim_{x\to a}f(x)+\lim_{x\to a}g(x)} {\displaystyle\lim_{x\to a}h(x)} \\ &= \boxed{\frac{L+M}{N}}. \end{aligned}$$

The condition $N\neq0$ is required for the quotient law.

2. Combine several limit laws

Suppose

$$\lim_{x\to2}f(x)=3 \qquad\text{and}\qquad \lim_{x\to2}g(x)=-1.$$

Evaluate

$$\lim_{x\to2}[2f(x)^2-3f(x)g(x)+g(x)].$$

Solution

Separate the expression using the sum, difference, constant-multiple, product, and power laws:

$$\begin{aligned} &\lim_{x\to2}[2f(x)^2-3f(x)g(x)+g(x)] \\ &= 2\left(\lim_{x\to2}f(x)\right)^2 - 3\left(\lim_{x\to2}f(x)\right) \left(\lim_{x\to2}g(x)\right) + \lim_{x\to2}g(x) \\ &= 2(3)^2-3(3)(-1)+(-1) \\ &= 18+9-1 \\ &= \boxed{26}. \end{aligned}$$

3. Prove the sum law from the formal definition

Suppose

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

Use the formal epsilon-delta definition of a limit to prove that

$$\boxed{ \lim_{x\to a}[f(x)+g(x)] = L+M. }$$

Your proof should identify suitable epsilon-dependent conditions for both functions, choose a single $\delta$ that guarantees both conditions simultaneously, and justify the inequality used to combine the two bounds.

Solution

Let $\varepsilon>0$ be arbitrary.

Since

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

there exists $\delta_1>0$ such that

$$0<|x-a|<\delta_1 \Longrightarrow |f(x)-L|<\frac{\varepsilon}{2}.$$

Likewise, since

$$\lim_{x\to a}g(x)=M,$$

there exists $\delta_2>0$ such that

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

Choose

$$\delta=\min(\delta_1,\delta_2).$$

Now suppose

$$0<|x-a|<\delta.$$

Because $\delta\le\delta_1$ and $\delta\le\delta_2$, both epsilon-dependent bounds apply:

$$|f(x)-L|<\frac{\varepsilon}{2} \qquad\text{and}\qquad |g(x)-M|<\frac{\varepsilon}{2}.$$

Using the triangle inequality,

$$\begin{aligned} |[f(x)+g(x)]-(L+M)| &= |[f(x)-L]+[g(x)-M]|\\ &\le |f(x)-L|+|g(x)-M|\\ &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\ &= \varepsilon. \end{aligned}$$

Therefore,

$$0<|x-a|<\delta \Longrightarrow |[f(x)+g(x)]-(L+M)|<\varepsilon.$$

By the formal definition of a limit,

$$\boxed{ \lim_{x\to a}[f(x)+g(x)] = \lim_{x\to a}f(x) + \lim_{x\to a}g(x) = L+M. }$$

4. Simplify by factoring

Evaluate

$$\lim_{h\to0}\frac{(3+h)^2-9}{h}.$$

Before opening the solution, identify the result of direct substitution and explain why another algebraic step is required.

Solution

Direct substitution produces the indeterminate form $\frac{0}{0}$. Expand and factor the numerator:

$$\begin{aligned} (3+h)^2-9 &= 9+6h+h^2-9 \\ &= 6h+h^2 \\ &= h(6+h). \end{aligned}$$

Therefore, for $h\neq0$,

$$\frac{(3+h)^2-9}{h} = \frac{h(6+h)}{h} = 6+h.$$

Now calculate the limit:

$$\begin{aligned} \lim_{h\to0}\frac{(3+h)^2-9}{h} &= \lim_{h\to0}(6+h) \\ &= 6+\lim_{h\to0}h \\ &= \boxed{6}. \end{aligned}$$

5. Rationalize with a conjugate

Evaluate

$$\lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}.$$

Identify why direct substitution is inconclusive and show every algebraic step.

Solution

Direct substitution produces $\frac{0}{0}$. Multiply by the conjugate:

$$\begin{aligned} \frac{\sqrt{x+5}-3}{x-4} &\cdot \frac{\sqrt{x+5}+3}{\sqrt{x+5}+3} \\ &= \frac{(x+5)-9} {(x-4)(\sqrt{x+5}+3)} \\ &= \frac{x-4} {(x-4)(\sqrt{x+5}+3)}. \end{aligned}$$

For $x\neq4$,

$$\frac{x-4} {(x-4)(\sqrt{x+5}+3)} = \frac1{\sqrt{x+5}+3}.$$

Therefore,

$$\begin{aligned} \lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4} &= \lim_{x\to4}\frac1{\sqrt{x+5}+3} \\ &= \frac1{\sqrt9+3} \\ &= \boxed{\frac16}. \end{aligned}$$

6. Decide whether the quotient law applies

Suppose

$$\lim_{x\to a}f(x)=4 \qquad\text{and}\qquad \lim_{x\to a}g(x)=0.$$

Can the quotient law be used to conclude that

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

has a particular value? Explain.

Solution

No. The quotient law requires the denominator limit to be nonzero:

$$\lim_{x\to a}g(x)\neq0.$$

Here,

$$\lim_{x\to a}g(x)=0,$$

so the quotient law does not apply. More information about $g(x)$ near $a$ is required to determine the behavior of the quotient.