Calculus I · Limits and Continuity · lesson

Limits With Complex Fractions

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 $3(3+h)$ :

\begin{aligned} \frac1{3+h}-\frac13 &=\frac3{3(3+h)}-\frac{3+h}{3(3+h)}\\ &=\frac{3-(3+h)}{3(3+h)}\\ &=\frac{-h}{3(3+h)}. \end{aligned}

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}.

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.

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.