Why Cancellation Requires More Terms
Some limits appear to have the form \(0/0\) because several terms in both the numerator and denominator cancel. In those cases, the first terms of the familiar Taylor expansions may tell us nothing about the quotient. We must continue until each expression has a nonzero term, then compare the powers and coefficients of those first surviving terms.
This is not a matter of expanding as far as possible. The goal is to expand far enough to determine the leading behavior. Once that behavior is known, a short factoring argument gives the limit and explains why the calculation works.
First Surviving Terms
Write \(h=x-a\) when considering a limit as \(x\) approaches \(a\). If an expression has its first nonzero Taylor term equal to \(Ch^m\), where \(C\ne 0\), then near \(a\) it behaves like \(Ch^m\). More precisely, under the appropriate Taylor expansion, it can be written as \(Ch^m+o(h^m)\). Here \(o(h^m)\) denotes a remainder whose quotient by \(h^m\) tends to zero as \(h\) tends to zero.
For example, if the numerator begins with \(Ah^4\) and the denominator with \(Bh^4\), their quotient approaches \(A/B\), provided \(B\ne 0\) and the denominator is nonzero sufficiently close to \(a\), except possibly at \(a\) itself. If the powers differ, the power of \(h\) left after cancellation determines whether the quotient tends to zero or fails to have a finite limit.
The Quotient Rule for Leading Taylor Terms
The following result makes that comparison precise. It applies whenever the stated local expansions hold; Taylor's theorem supplies such expansions when the functions have sufficient derivatives near the point. For standard functions represented by convergent Taylor series, the series can be used directly within their interval of convergence.
- If \(m=n\), \(\displaystyle\lim_{h\to0}\frac{f(a+h)}{g(a+h)}=\frac{A}{B}\).
- If \(m>n\), \(\displaystyle\lim_{h\to0}\frac{f(a+h)}{g(a+h)}=0\).
- If \(m<n\), the quotient has no finite limit; its magnitude grows without bound, though its sign may differ on the two sides.
Why the Comparison Works
The proof is a factorization of the dominant powers. By the definition of little-oh notation, there are functions \(\varepsilon(h)\) and \(\delta(h)\) that tend to zero as \(h\to0\) and satisfy
For sufficiently small \(h\), \(B+\delta(h)\) is nonzero, because it approaches the nonzero number \(B\). Thus, for nonzero \(h\) in that neighborhood,
The second factor approaches \(A/B\), a finite nonzero number. If \(m=n\), the power factor is \(1\), so the quotient approaches \(A/B\). If \(m>n\), the factor \(h^{m-n}\) tends to zero, so the whole quotient tends to zero. If \(m<n\), its absolute value grows without bound because the nonzero second factor approaches \(A/B\), while \(|h|^{m-n}\) grows without bound. The sign can depend on the parity of \(m-n\) and on which side of zero is considered, so a two-sided infinite limit is not guaranteed.
Worked Example: Equal Orders After Several Cancellations
Worked Example: Equal Orders After Several Cancellations
Evaluate \(\displaystyle\lim_{x\to0}\frac{e^x-1-x-\frac{x^2}{2}-\frac{x^3}{6}}{1-\cos x-\frac{x^2}{2}}\).
In the numerator, the constant and the terms through degree three have been subtracted. Expand through the first remaining term:
so the numerator is \(\frac{x^4}{24}+O(x^5)\). For the denominator,
and therefore
Both expressions have order four. Factor out \(x^4\) from each:
The denominator of this last quotient approaches \(-1/24\), which is nonzero. Taking the limit gives
Check: The quotient of the leading terms is \((x^4/24)/(-x^4/24)=-1\) for \(x\ne0\). The factored remainder terms approach those same nonzero coefficients, confirming that no higher-order terms change the limit.
Worked Example: The Numerator Has Higher Order
Worked Example: The Numerator Has Higher Order
Evaluate \(\displaystyle\lim_{x\to0}\frac{\sin x-x+\frac{x^3}{6}-\frac{x^5}{120}}{1-\cos x-\frac{x^2}{2}+\frac{x^4}{24}}\).
Use the sine expansion through degree seven. After the displayed cancellations, the first surviving term is degree seven:
For the denominator, use the cosine expansion through degree six:
which gives
The numerator has order seven and the denominator order six. Factoring their leading powers yields
The fraction after \(x\) approaches a finite number, since its denominator approaches \(1/720\ne0\), while \(x\to0\). Therefore,
Check: The leading-term quotient is \(\left(-x^7/5040\right)/(x^6/720)=-x/7\), which tends to zero. This agrees with the limit obtained after including the remainders.
Worked Example: Comparing Two Different Taylor Expansions
Worked Example: Comparing Two Different Taylor Expansions
Evaluate \(\displaystyle\lim_{x\to0}\frac{\ln(1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}}{e^x-1-x-\frac{x^2}{2}-\frac{x^3}{6}}\).
The logarithm expansion near zero begins
so the numerator simplifies to
Expand the exponential through degree four as well:
and obtain
The orders match, so factor \(x^4\) and compare the coefficients:
Check: The ratio of the first surviving terms is \((-x^4/4)/(x^4/24)=-6\) for \(x\ne0\). Both factored expressions approach nonzero coefficients, so the higher-order remainders do not alter the limit.
A Common Mistake: Stopping Before Anything Survives
A truncation is useful only if it retains the first nonzero term needed for the limit. For instance, in the first worked example, using Taylor polynomials only through degree three makes both the numerator and denominator appear to be zero. That does not show that the original quotient is zero, nor does it determine its limit. It shows only that the chosen truncations were too short.
Another error is to compare coefficients before checking the powers. If the leading powers differ, the quotient includes a remaining power of \(x-a\); coefficient division alone then gives the wrong conclusion. Always record both the first surviving power and its coefficient.
A Reliable Procedure
Set \(h=x-a\) when the limit is as \(x\to a\), so the expansion is organized in powers of the quantity approaching zero.
For the numerator and denominator separately, subtract the specified terms and find the first term that remains nonzero.
Include the remainder at an order that confirms the claimed leading term, rather than stopping at a truncation that vanishes entirely.
Equal powers give a coefficient ratio; a higher numerator power gives zero; a lower numerator power prevents a finite limit.
Confirm that the denominator's leading coefficient is nonzero and that the original denominator is nonzero for nearby points where the quotient is evaluated.
Check Your Understanding
For each question, identify the first surviving terms before deciding what the quotient does.
- If a numerator begins with \(5h^6+o(h^6)\) and a denominator begins with \(-2h^6+o(h^6)\), what is the quotient limit as \(h\to0\)?
- If the numerator begins with \(3h^8+o(h^8)\) and the denominator with \(4h^5+o(h^5)\), what is the limit?
- Why does a degree-three Taylor truncation that makes both numerator and denominator zero fail to determine a quotient limit?
- What additional condition on the denominator is needed to form the quotient near, but not necessarily at, the limit point?
- If the numerator and denominator have different leading powers, why is dividing only their leading coefficients insufficient?
