Calculus I · Limits and Continuity · lesson

Direct Substitution for Limits

Concept

Learning objectives

Evaluate limits of familiar continuous functions by substitution; interpret the result of substitution before choosing a more complicated method.

Evaluating Finite Limits

Start With Direct Substitution

Once you understand what a limit means, most homework problems become a method-selection problem. The first move should almost always be the simplest one:

Method

First Move for Nearly Every Limit

Substitute the target input into the expression.

• If you get an ordinary real number, that is usually the limit. • If you get $0/0$ , simplify the expression and try again. • If you get a nonzero number divided by zero, analyze one-sided infinite behavior. • If the input approaches infinity and you get an indeterminate form such as $\infty/\infty$ , compare dominant terms.

Why does substitution work for so many functions? Because polynomials, rational functions away from zero denominators, root functions on their domains, and trigonometric functions on their domains are continuous. Continuity will be developed carefully in Section 5. For now, think of a continuous graph as one with no break at the target point.

Guided walkthrough

Substitute and stop

Evaluate

\lim_{x\to2}(x+5).

Show worked solution

Replace $x$ by $2$ :

2+5=7.

Nothing breaks, no denominator becomes zero, and no special method is needed. Therefore,

\boxed{\lim_{x\to2}(x+5)=7}.

Interactive checkdirect-sub-01

Evaluate $\lim_{x\to2}(3x+1)$ .

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Show hint

Substitute because the linear function is continuous.

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Worked example

Polynomial substitution

Evaluate

\lim_{x\to-2}(3x^3-4x^2+5x-1).

Show worked solution

Polynomials are continuous everywhere, so substitute $x=-2$ :

\begin{aligned} 3(-2)^3-4(-2)^2+5(-2)-1 &=3(-8)-4(4)-10-1\\ &=-24-16-10-1\\ &=\boxed{-51}. \end{aligned}

Worked example

Rational function with a safe denominator

Evaluate

\lim_{x\to3}\frac{x^2+2x-1}{x+4}.

Show worked solution

Substitute first:

\frac{3^2+2(3)-1}{3+4} =\frac{9+6-1}{7} =\frac{14}{7} =\boxed{2}.

The denominator approaches $7$ , not zero, so the quotient law is valid.

Worked example

A root function

Evaluate

\lim_{x\to5}\sqrt{3x+1}.

Show worked solution

The expression under the root approaches

3(5)+1=16.

Therefore,

\lim_{x\to5}\sqrt{3x+1}=\sqrt{16}=\boxed{4}.

Common mistake

Students sometimes overcomplicate a limit because the section is about limits. If substitution gives a legal real number, do not factor six polynomials, draw a sign chart, or summon l'Hospital's Rule from a future section. Stop.

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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