Calculus I · Limits and Continuity · lesson

Evaluating Radical Limits With Conjugates

Concept

Learning objectives

Use a conjugate to remove a radical difference that creates $0/0$ , then evaluate the simplified limit.

Rationalizing Radicals

A conjugate changes the sign between two terms:

\sqrt{A}-B \qquad\longleftrightarrow\qquad \sqrt{A}+B.

Multiplying conjugates uses the difference-of-squares identity:

(u-v)(u+v)=u^2-v^2.

This removes the radical difference.

Guided walkthrough

One conjugate, every line shown

Evaluate

\lim_{x\to0}\frac{\sqrt{x+4}-2}{x}.

Show worked solution

Direct substitution gives

\frac{\sqrt4-2}{0}=\frac00.

Multiply the fraction by $1$ written as the conjugate over itself:

\frac{\sqrt{x+4}-2}{x} \cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}.

Multiply the numerator using difference of squares:

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

Therefore,

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

For $x\ne0$ , cancel $x$ :

\frac{1}{\sqrt{x+4}+2}.

Now substitute $x=0$ :

\frac1{\sqrt4+2}=\frac1{2+2}=\boxed{\frac14}.

Interactive checkconjugate-flow-01

Evaluate $\lim_{x\to0}\frac{\sqrt{x+4}-2}{x}$ .

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Show hint

Use the conjugate $\sqrt{x+4}+2$ .

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Worked example

Radical in the denominator

Evaluate

\lim_{x\to9}\frac{x-9}{\sqrt{x}-3}.

Show worked solution

Substitution gives $0/0$ . Notice that

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

Therefore, for $x\ne9$ ,

\frac{x-9}{\sqrt{x}-3}=\sqrt{x}+3.

Now substitute:

\sqrt9+3=3+3=\boxed{6}.

You could also multiply by the conjugate. Recognizing the hidden difference of squares is simply faster.

Worked example

Exam-level: two radical terms

Evaluate

\lim_{x\to0}\frac{\sqrt{1+3x}-\sqrt{1-x}}{x}.

Show worked solution

The numerator is a difference of radicals, so use its conjugate:

\begin{aligned} &\frac{\sqrt{1+3x}-\sqrt{1-x}}{x} \cdot \frac{\sqrt{1+3x}+\sqrt{1-x}}{\sqrt{1+3x}+\sqrt{1-x}}\\ &=\frac{(1+3x)-(1-x)}{x\left(\sqrt{1+3x}+\sqrt{1-x}\right)}\\ &=\frac{4x}{x\left(\sqrt{1+3x}+\sqrt{1-x}\right)}\\ &=\frac4{\sqrt{1+3x}+\sqrt{1-x}},\qquad x\ne0. \end{aligned}

Now substitute:

\frac4{1+1}=\boxed{2}.

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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