Calculus I · Limits and Continuity · lesson

Limit Laws and How to Use Them

Concept

Learning objectives

Use sums, differences, constant multiples, products, quotients, powers, and roots to build limits from simpler limits.

Limit Laws

Suppose

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

The basic limit laws are:

Theorem

Limit Laws

\begin{aligned} \lim_{x\to a}[f(x)+g(x)]&=L+M,\\ \lim_{x\to a}[f(x)-g(x)]&=L-M,\\ \lim_{x\to a}[cf(x)]&=cL,\\ \lim_{x\to a}[f(x)g(x)]&=LM,\\ \lim_{x\to a}\frac{f(x)}{g(x)}&=\frac{L}{M},\qquad M\ne0,\\ \lim_{x\to a}[f(x)]^n&=L^n. \end{aligned}

When the relevant root is defined,

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

Concept

The laws say that limits cooperate with ordinary arithmetic, provided the arithmetic itself remains legal. If one part approaches $3$ and another approaches $5$ , their sum approaches $8$ . The only major warning in the basic laws is division: the denominator's limit may not be zero.

Guided walkthrough

Building a limit from known pieces

Suppose

\lim_{x\to2}f(x)=3 \qquad\text{and}\qquad \lim_{x\to2}g(x)=5.

Find

\lim_{x\to2}[2f(x)+g(x)].

Show worked solution

The constant multiple law gives

\lim_{x\to2}2f(x)=2(3)=6.

Then the sum law gives

6+5=11.

Therefore,

\boxed{\lim_{x\to2}[2f(x)+g(x)]=11}.

Worked example

Several laws at once

Suppose

\lim_{x\to1}f(x)=-2, \qquad \lim_{x\to1}g(x)=4.

Evaluate

\lim_{x\to1}\frac{[f(x)]^2+3g(x)}{g(x)-f(x)}.

Show worked solution

Apply the laws to each part:

[f(x)]^2\to(-2)^2=4,

3g(x)\to3(4)=12,

so the numerator approaches

4+12=16.

The denominator approaches

4-(-2)=6.

Because $6\ne0$ , the quotient law applies:

\boxed{\frac{16}{6}=\frac83}.

Quick check

If $f(x)\to3$ and $g(x)\to-1$ as $x\to4$ , evaluate

\lim_{x\to4}[f(x)g(x)+2f(x)].

Answer. $3(-1)+2(3)=3$ .

After the explanation

Use the section idea

Reading lens

What did direct substitution reveal, and which algebraic move removes the obstacle without changing nearby behavior?

Mental model

Substitution is a diagnostic first move: a real number usually finishes the problem, while an indeterminate form asks for a structural rewrite.

Decision

Match the obstacle to the algebra—factor polynomial zeros, rationalize radicals, combine complex fractions, and split absolute values into one-sided cases.

Common trap

Zero over zero is not an answer, and cancellation is legal only for factors after the expression has been rewritten as a product.

Check yourself

You are ready to move on when you can justify why each rewrite preserves nearby values even if the original expression is undefined at the target.

Source & rights

Original instruction with traceable references.

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