Calculus I · Limits and Continuity · lesson

Limits With Complex Fractions

Learn Limits With Complex Fractions with plain-language explanations, guided examples, worked homework methods, interactive checks, and exam-style practice.

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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 Limits With Complex Fractions; the worked mathematics is evidence for the idea, not a substitute for it.

This page connects Evaluating Radical Limits With Conjugates to Limits With Absolute Values and Piecewise Functions. Read the explanation first, predict each example’s next move, and only then compare the written solution.

Concept

Learning objectives

Combine smaller fractions using a common denominator before simplifying a limit whose numerator or denominator contains fractions.

Complex Fractions

Concept

A complex fraction is just a large fraction containing smaller fractions. Do not try to "cancel through" addition. First turn the top or bottom into one ordinary fraction. Then divide.

Guided walkthrough

Combine the numerator first

Evaluate

\lim_{x\to0}\frac{\frac1{x+1}-1}{x}.

Show worked solution

Direct substitution gives $0/0$ . Combine the two terms in the numerator:

\frac1{x+1}-1 =\frac1{x+1}-\frac{x+1}{x+1} =\frac{1-(x+1)}{x+1} =\frac{-x}{x+1}.

Now place that result over $x$ :

\frac{\frac{-x}{x+1}}{x} =\frac{-x}{x+1}\cdot\frac1x =-\frac1{x+1},\qquad x\ne0.

Substitute:

-\frac1{0+1}=\boxed{-1}.

Worked example

Difference quotient for a reciprocal

Evaluate

\lim_{h\to0} \frac{\frac1{3+h}-\frac13}{h}.

Show worked solution

Combine the fractions in the numerator using denominator $3(3+h)$ :

\frac1{3+h}-\frac13 \quad =\frac3{3(3+h)}-\frac{3+h}{3(3+h)}\quad \quad =\frac{3-(3+h)}{3(3+h)}\quad \quad =\frac{-h}{3(3+h)}.

Now divide by $h$ :

\frac{\frac{-h}{3(3+h)}}{h} =\frac{-h}{3(3+h)}\cdot\frac1h =-\frac1{3(3+h)}.

Take the limit:

-\frac1{3(3)}=\boxed{-\frac19}.

Worked example

Exam-level: two rational terms

Evaluate

\lim_{x\to2} \frac{\frac1x-\frac12}{x-2}.

Show worked solution

Combine the numerator:

\frac1x-\frac12=\frac{2-x}{2x}=-\frac{x-2}{2x}.

Therefore,

\frac{-\frac{x-2}{2x}}{x-2} =-\frac1{2x},\qquad x\ne2.

Now substitute:

-\frac1{2(2)}=\boxed{-\frac14}.

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.

Complex Fractions

Concept

A complex fraction is just a large fraction containing smaller fractions. Do not try to "cancel through" addition. First turn the top or bottom into one ordinary fraction. Then divide.

Guided walkthrough

Combine the numerator first

Evaluate

\lim_{x\to0}\frac{\frac1{x+1}-1}{x}.

Show worked solution

Direct substitution gives $0/0$ . Combine the two terms in the numerator:

\frac1{x+1}-1 =\frac1{x+1}-\frac{x+1}{x+1} =\frac{1-(x+1)}{x+1} =\frac{-x}{x+1}.

Now place that result over $x$ :

\frac{\frac{-x}{x+1}}{x} =\frac{-x}{x+1}\cdot\frac1x =-\frac1{x+1},\qquad x\ne0.

Substitute:

-\frac1{0+1}=\boxed{-1}.

Worked example

Difference quotient for a reciprocal

Evaluate

\lim_{h\to0} \frac{\frac1{3+h}-\frac13}{h}.

Show worked solution

Combine the fractions in the numerator using denominator $3(3+h)$ :

\frac1{3+h}-\frac13 \quad =\frac3{3(3+h)}-\frac{3+h}{3(3+h)}\quad \quad =\frac{3-(3+h)}{3(3+h)}\quad \quad =\frac{-h}{3(3+h)}.

Now divide by $h$ :

\frac{\frac{-h}{3(3+h)}}{h} =\frac{-h}{3(3+h)}\cdot\frac1h =-\frac1{3(3+h)}.

Take the limit:

-\frac1{3(3)}=\boxed{-\frac19}.

Worked example

Exam-level: two rational terms

Evaluate

\lim_{x\to2} \frac{\frac1x-\frac12}{x-2}.

Show worked solution

Combine the numerator:

\frac1x-\frac12=\frac{2-x}{2x}=-\frac{x-2}{2x}.

Therefore,

\frac{-\frac{x-2}{2x}}{x-2} =-\frac1{2x},\qquad x\ne2.

Now substitute:

-\frac1{2(2)}=\boxed{-\frac14}.

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

Complex Fractions

Related resources

Original instruction with traceable references.

Chapter 2: Finite limits and algebra

Complex Fractions

Related resources

Original instruction with traceable references.