Calculus I · Limits and Continuity · lesson

Evaluating Radical Limits With Conjugates

Learn Evaluating Radical Limits With Conjugates with plain-language explanations, guided examples, worked homework methods, interactive checks, and exam-style p

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Where this chapter fits

Chapter 2: Finite limits and algebra

Turn indeterminate forms into solvable expressions using substitution, limit laws, factoring, conjugates, and piecewise reasoning.

Reading lens: What did direct substitution reveal, and which algebraic move removes the obstacle without changing nearby behavior? Keep that question in view while reading Evaluating Radical Limits With Conjugates; the worked mathematics is evidence for the idea, not a substitute for it.

This page connects Evaluating Limits by Factoring to Limits With Complex Fractions. Read the explanation first, predict each example’s next move, and only then compare the written solution.

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

Your work stays on this device. No account or AI grader is used.

Show hint

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

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

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:

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

Now substitute:

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

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

The verified handoff declares original composition and requires owner provenance review. BetterGrades-original material remains separate from public-domain references; no source textbook PDF is published here.

Chapter 2: Finite limits and algebra

Turn indeterminate forms into solvable expressions using substitution, limit laws, factoring, conjugates, and piecewise reasoning.

This page connects Evaluating Limits by Factoring to Limits With Complex Fractions. Read the explanation first, predict each example’s next move, and only then compare the written solution.

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

Your work stays on this device. No account or AI grader is used.

Show hint

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

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

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:

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

Now substitute:

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

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

Chapter 2: Finite limits and algebra

Rationalizing Radicals

Related resources

Original instruction with traceable references.

Chapter 2: Finite limits and algebra

Rationalizing Radicals

Related resources

Original instruction with traceable references.