Taylor Expansions for Indeterminate Limits

Why Taylor Expansions Help with Indeterminate Limits

When a numerator and denominator both approach zero, direct substitution gives the form \(0/0\), which does not determine the limit. The expressions may approach zero at different rates: one may contain a term proportional to \(x-a\), while another first becomes nonzero at \((x-a)^3\). Those rates determine the quotient.

Taylor expansions make the rates visible. The essential task is to expand far enough to find the first nonzero term in both numerator and denominator. Terms that cancel must not be mistaken for the leading term.

This method is for limits near a finite point where the functions admit the required Taylor expansions. It does not by itself evaluate every indeterminate form, such as a limit at infinity.

Local Taylor Expansions and Vanishing Order

For a limit as \(x\) approaches \(a\), write \(h=x-a\). Then \(h\) approaches zero, and the expansion is centered at \(a\). If \(f\) has continuous derivatives through order \(k\) near \(a\), its Taylor expansion through degree \(k\) can be written

f(a+h)=f(a)+f'(a)h+\frac{f''(a)}{2!}h^2+\cdots+ \frac{f^{(k)}(a)}{k!}h^k+o(h^k).

Here \(r(h)=o(h^k)\) means that \(r(h)/h^k\) tends to zero as \(h\) tends to zero. Such a remainder is smaller than \(h^k\) in the limiting sense. The notation \(O(h^k)\), used occasionally, means that the absolute value of the remainder is at most a constant times \(|h|^k\) sufficiently near zero.

If \(f(a)=0\) and the first nonzero Taylor coefficient is the coefficient of \(h^m\), then \(f(a+h)\) starts with a nonzero multiple of \(h^m\). We say that \(f\) vanishes to order \(m\) at \(a\). This description concerns the first nonzero term; terms after it may also be present.

Definition: If \(f(a)=f'(a)=\cdots=f^{(m-1)}(a)=0\) and \(f^{(m)}(a)\ne 0\), then \(f\) vanishes to order \(m\) at \(a\). Its leading Taylor term is \(\frac{f^{(m)}(a)}{m!}(x-a)^m\).

The Leading-Term Quotient Rule

The order of vanishing gives a general test for a quotient. Suppose the numerator \(F\) and denominator \(G\) have Taylor expansions about \(a\) whose first nonzero terms are \(A(x-a)^m\) and \(B(x-a)^n\), respectively, where \(A\ne 0\) and \(B\ne 0\). Then, setting \(h=x-a\),

F(a+h)=h^m(A+\varepsilon(h)),\qquad G(a+h)=h^n(B+\delta(h)),

where \(\varepsilon(h)\to 0\) and \(\delta(h)\to 0\). Since \(B\ne 0\), the factor \(B+\delta(h)\) stays nonzero for all sufficiently small \(h\). For \(h\ne 0\) close to zero, division therefore gives

\frac{F(a+h)}{G(a+h)} =h^{m-n}\frac{A+\varepsilon(h)}{B+\delta(h)}.

Theorem: Under these assumptions, if \(m=n\), then \(\displaystyle\lim_{x\to a}\frac{F(x)}{G(x)}=\frac{A}{B}\). If \(m>n\), the quotient tends to zero. If \(m<n\), the quotient is unbounded in magnitude near \(a\), so there is no finite two-sided limit. Whether it tends to positive or negative infinity, or has different one-sided behavior, depends on the signs and parity of the powers.

This rule follows because the second factor in the quotient tends to \(A/B\), a finite nonzero number. If \(m=n\), the power of \(h\) is zero. If \(m>n\), the remaining power \(h^{m-n}\) tends to zero. If \(m<n\), its absolute value grows without bound as \(h\) approaches zero. This is why finding the first nonzero terms—not merely writing several terms—is the central step.

A Reliable Expansion Procedure

Shift to the limiting point.
Set \(h=x-a\), so that the limit becomes a limit as \(h\to 0\).

Expand each part.
Use Taylor expansions centered at \(a\), keeping terms through the first nonzero term in each expression.

Account for cancellation.
Combine the terms in the numerator and denominator before deciding their orders of vanishing.

Compare leading powers.
Factor those powers and take the limit of the remaining quotient.

Check the result.
Verify that the retained terms really are the first nonzero ones and that the discarded remainders cannot change the limit.

Worked Examples

Worked Example: An Exponential Remainder

Evaluate \(\displaystyle\lim_{x\to 0}\frac{e^x-1-x}{x^2}\).

Both numerator and denominator approach zero. The constant and linear terms of \(e^x\) cancel in the numerator, so expand through the quadratic term:

e^x=1+x+\frac{x^2}{2}+o(x^2).

Subtracting \(1+x\) gives

e^x-1-x=\frac{x^2}{2}+o(x^2).

For \(x\ne 0\), divide by \(x^2\):

\frac{e^x-1-x}{x^2} =\frac{1}{2}+\frac{o(x^2)}{x^2} =\frac{1}{2}+o(1).

Since \(o(1)\) tends to zero, the limit is \(1/2\). The check is that the numerator’s first nonzero term is \(x^2/2\), matching the denominator’s \(x^2\); the ratio of their coefficients is \((1/2)/1=1/2\).

Worked Example: Cancellation Between Trigonometric Terms

Evaluate \(\displaystyle\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}\).

Expand each function far enough to see the cubic terms. The needed expansions are

\sin x=x-\frac{x^3}{6}+o(x^3), \qquad \cos x=1-\frac{x^2}{2}+o(x^2).

Multiplying the second expansion by \(x\) gives \(x\cos x=x-\frac{x^3}{2}+o(x^3)\). Subtract this from the expansion for \(\sin x\):

\begin{aligned} \sin x-x\cos x &=\left(x-\frac{x^3}{6}+o(x^3)\right) -\left(x-\frac{x^3}{2}+o(x^3)\right)\\ &=\left(-\frac{1}{6}+\frac{1}{2}\right)x^3+o(x^3)\\ &=\frac{x^3}{3}+o(x^3). \end{aligned}

Thus, for \(x\ne 0\), the quotient is \(\frac{1}{3}+o(1)\), and the limit is \(1/3\). As a check, the linear terms cancel exactly, while the cubic coefficient is \(-1/6+1/2=1/3\), so division by \(x^3\) leaves the stated value.

Worked Example: A Logarithmic Numerator

Evaluate \(\displaystyle\lim_{x\to 0}\frac{\ln(1+x)-x+x^2/2}{x^3}\).

The constant, linear, and quadratic terms cancel, so the logarithm must be expanded through degree three:

\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3).

Substitute this expansion into the numerator and combine like terms:

\begin{aligned} \ln(1+x)-x+\frac{x^2}{2} &=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)-x+\frac{x^2}{2}\\ &=\frac{x^3}{3}+o(x^3). \end{aligned}

After division by \(x^3\), the quotient is \(\frac{1}{3}+o(1)\), so the limit is \(1/3\). This checks against the leading-term rule: both expressions vanish to order three, and their leading coefficients are \(1/3\) and \(1\).

Worked Example: Expanding About a Nonzero Point

Evaluate \(\displaystyle\lim_{x\to 1}\frac{e^x-e-e(x-1)}{(x-1)^2}\).

Set \(h=x-1\), so \(x=1+h\) and \(h\to 0\). Expand the exponential about \(1\), rather than about zero:

e^{1+h}=e\left(1+h+\frac{h^2}{2}+o(h^2)\right) =e+eh+\frac{e}{2}h^2+o(h^2).

The numerator is \(e^{1+h}-e-eh\). Substitution and cancellation yield

e^{1+h}-e-eh =e+eh+\frac{e}{2}h^2+o(h^2)-e-eh =\frac{e}{2}h^2+o(h^2).

The denominator is \(h^2\), so the quotient is \(\frac{e}{2}+o(1)\) and the limit is \(e/2\). To check, the numerator and denominator both have order two in \(h\); dividing their leading coefficients gives \((e/2)/1=e/2\).

How Many Terms Are Enough?

The appropriate expansion degree is determined by cancellation, not by a fixed rule such as “always keep three terms.” If the numerator’s constant, linear, and quadratic contributions cancel, terms only through degree two leave no information about its leading behavior. You must continue until a nonzero term appears. The denominator must be treated the same way.

In practice, the denominator’s first nonzero term often gives a useful target. If it is proportional to \((x-a)^n\), the numerator must be expanded at least through degree \(n\) to test whether its first surviving term has the same order, a higher order, or a lower order. If all terms through degree \(n\) cancel in the numerator, then expand further: the numerator might still have a nonzero term at a higher order.

The remainder matters too. An expansion that displays a candidate leading term must also have a remainder small enough not to overwhelm it. For example, to claim \(F(a+h)=Ah^m+o(h^m)\) with \(A\ne 0\), the remainder divided by \(h^m\) must tend to zero. That condition guarantees that \(Ah^m\) really determines the leading behavior near the point.

Common Pitfall: Stopping Before the First Surviving Term

For the first example, a tempting but incorrect calculation is to use only \(e^x\approx 1+x\), conclude that \(e^x-1-x\approx 0\), and then treat the quotient as zero. This is not a valid conclusion: the approximation has discarded the very term that controls the quotient.

Keeping the next term corrects the calculation:

e^x=1+x+\frac{x^2}{2}+o(x^2) \quad\Longrightarrow\quad \frac{e^x-1-x}{x^2} =\frac{1}{2}+o(1) \longrightarrow\frac{1}{2}.

A related error is to cancel powers before combining the expressions in the numerator. In \(\sin x-x\cos x\), each term has a linear contribution, but those contributions cancel. The numerator’s order is therefore not determined by looking at either term in isolation. Combine first, identify the first nonzero coefficient, and only then compare powers.

When the Method Does Not Give a Finite Limit

The leading-term comparison can rule out a finite limit as well as evaluate one. If the numerator vanishes to order one and the denominator to order two, the quotient contains a factor proportional to \(1/(x-a)\). Its magnitude becomes unbounded, and its signs on the two sides may differ. If the first nonzero terms cannot be established—for example, because the available expansion or its remainder is insufficient—then the comparison is inconclusive. More terms or another valid method are needed.

The method is valuable because it turns a \(0/0\) expression into a comparison of local rates. It also explains why apparently complicated limits can have simple answers: once the cancellations are exposed, only the first surviving coefficients may matter.

Check Your Understanding

For each question, identify the first nonzero terms before deciding what information they provide.

In \(\displaystyle\frac{\sin x-x}{x^3}\) as \(x\to 0\), through what degree must the sine expansion be taken to find the first surviving numerator term?
If the numerator has leading term \(5(x-a)^3\) and the denominator has leading term \(2(x-a)^3\), what does the leading-term rule predict for the limit?
If the numerator has leading term \(4(x-a)^5\) and the denominator has leading term \(-2(x-a)^2\), what is the finite limit?
Why is an expansion with remainder \(o(h^2)\) sufficient to establish a nonzero quadratic leading term, but an expansion with an unspecified remainder may not be?
What must you check before claiming that the first terms written for a numerator determine its order of vanishing?