AcademyPower Series
Questions

Academy

Taylor Limits

Level 1 - Math I (Physics) topic page in Power Series.

Principle

Taylor expansions make many limits simpler by replacing functions with their leading non-zero terms. Near the limiting point, the lowest power that survives usually controls the limit.

Leading-term idea
\[f(x)=a_k(x-c)^k+\text{higher powers}\]

This is especially useful for limits that give \(0/0\). Instead of repeatedly differentiating, expand the numerator and denominator until the first non-zero terms appear.

Notation

\(c\)
point approached in the limit
\(x-c\)
small quantity tending to zero
\(a_k(x-c)^k\)
first non-zero term in an expansion
\(O((x-c)^m)\)
terms of order m or higher
\(\sim\)
has the same leading behaviour as

For Level 1 calculations, it is usually enough to write enough terms explicitly and then identify the first non-zero power.

Method

Step 1: Shift the limit point if needed

If \(x\to c\), use \(u=x-c\), so \(u\to0\). Expand in powers of \(u\).

Step 2: Expand each function only as far as needed

Use standard series such as

Useful small-x expansions
\[e^x=1+x+\frac{x^2}{2}+\cdots,\quad \sin x=x-\frac{x^3}{3!}+\cdots,\quad \cos x=1-\frac{x^2}{2}+\cdots\]

Step 3: Cancel the lowest common power

After substituting expansions, factor out the smallest power that appears in both numerator and denominator.

Typical form
\[\frac{a_kx^k+\text{higher powers}}{b_kx^k+\text{higher powers}}\]
Factor the leading power
\[\frac{x^k(a_k+\text{smaller corrections})}{x^k(b_k+\text{smaller corrections})}\]
Cancel and take the limit
\[\lim_{x\to0}\frac{a_k+\text{smaller corrections}}{b_k+\text{smaller corrections}}=\frac{a_k}{b_k}\]

Rules

Sine leading term
\[\sin x\sim x\quad(x\to0)\]
Cosine difference leading term
\[1-\cos x\sim\frac{x^2}{2}\quad(x\to0)\]
Exponential difference leading term
\[e^x-1\sim x\quad(x\to0)\]
Logarithm leading term
\[\ln(1+x)\sim x\quad(x\to0)\]

These leading behaviours assume radians for trigonometric functions.

Examples

Question
Evaluate
\[\lim_{x\to0}\frac{1-\cos x}{x^2}\]
Answer
Use
\[\cos x=1-x^2/2+\cdots\]
Then
\[1-\cos x=x^2/2+\cdots\]
Divide by \(x^2\) to get
\[1/2+\cdots\]
so the limit is
\[1/2\]

Checks

  • Expand far enough to find the first non-zero term after cancellation.
  • Do not keep unnecessary higher powers once the leading terms decide the limit.
  • Trigonometric Taylor limits require radians.
  • If leading terms cancel, expand to the next order.

Previous

Remainders

Next

Linear Systems