Applications of Taylor's Theorem

Turning Taylor Approximations into Conclusions

Taylor’s Theorem gives more than an approximation with an error estimate. It can reveal a function’s sign near a point, determine whether a stationary point is a local extremum, and evaluate a limit after several lower-order terms cancel. The key is to identify the first Taylor term that does not vanish.

Throughout, let $a$ be a point in an open interval on which the functions under consideration have the derivatives required. Taylor’s Theorem and its remainder forms were established earlier in the course. We will also use Taylor’s Theorem, Peano Form: if $f$ has derivatives through order $m$ near $a$, then its degree-$m$ Taylor expansion can be written as a polynomial plus a remainder $o((x-a)^m)$. Here this notation means that the remainder, divided by $(x-a)^m$, tends to zero as $x\to a$, with $x\ne a$.

f(a+h)=\sum_{k=0}^{m}\frac{f^{(k)}(a)}{k!}h^k+o(h^m).

If the constant term and several subsequent coefficients vanish, the first nonzero term often controls the behavior. A remainder that is small compared with that term cannot change its sign sufficiently close to the center. The following result makes this principle precise.

Theorem (Local Sign from the First Nonzero Taylor Term): Let $m\geq1$, and suppose $f$ has derivatives through order $m$ on an open interval containing $a$. Assume $f(a)=0$, $f^{(k)}(a)=0$ for $1\leq k<m$, and $f^{(m)}(a)\ne0$. For all sufficiently small nonzero $h$, the sign of $f(a+h)$ is the sign of $$ \frac{f^{(m)}(a)}{m!}h^m. $$ In particular, if $m$ is even, $f(a)$ is a strict local minimum when $f^{(m)}(a)>0$, and a strict local maximum when $f^{(m)}(a)<0$. If $m$ is odd, $f$ takes opposite signs on the two sides of $a$, so $a$ is neither a local maximum nor a local minimum.

Proof. Set $c=f^{(m)}(a)/m!$, which is nonzero. Taylor’s Theorem, Peano Form and the vanishing hypotheses give

f(a+h)=ch^m+r(h), \qquad \lim_{h\to0}\frac{r(h)}{h^m}=0.

Choose $\delta>0$ small enough that whenever $0<|h|<\delta$,

\left|\frac{r(h)}{h^m}\right|<\frac{|c|}{2}.

Then $f(a+h)=h^m(c+r(h)/h^m)$. The factor in parentheses has the same sign as $c$: its distance from $c$ is less than $|c|/2$. Thus $f(a+h)$ has the sign of $ch^m$, proving the first claim. If $m$ is even, $h^m>0$ on both sides of zero, so the sign of $f(a+h)$ is the sign of $c$ on both sides. This gives the stated strict local minimum or maximum. If $m$ is odd, $h^m$ has opposite signs on the two sides, so $f(a+h)$ does too. Values on both sides then rule out either kind of local extremum. $\square$

Worked Example: A Higher-Order Strict Local Minimum

Consider $f(x)=x^4/(1+x^2)$ near $a=0$. The exact identity

\frac{x^4}{1+x^2}=x^4-\frac{x^6}{1+x^2}

shows that $f(x)=x^4+o(x^4)$: indeed, after dividing the second term by $x^4$, its absolute value is $x^2/(1+x^2)$, which tends to zero. The first nonzero Taylor term is therefore $x^4$, with positive coefficient. The local sign theorem shows that $f(0)=0$ is a strict local minimum. In this example $f$ is also visibly positive for $x\ne0$, but the expansion illustrates how the same conclusion can be reached from the leading Taylor term even when the sign is less apparent from the original formula.

Limits After Cancellation

Taylor expansions are especially useful for limits when the numerator and denominator both vanish. Direct substitution may give $0/0$, and subtracting a few low-order terms may still leave an indeterminate quotient. If both expressions have the same first nonzero order, their quotient approaches the ratio of the corresponding coefficients.

Theorem (Limit of a Quotient with Matching Vanishing Order): Let $m\geq1$. Suppose $f$ and $g$ have derivatives through order $m$ near $a$, and for each of them the value and derivatives of orders $1,\ldots,m-1$ vanish at $a$. If $f^{(m)}(a)\ne0$ and $g^{(m)}(a)\ne0$, then $g(x)\ne0$ for all sufficiently close $x\ne a$, and $$ \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{f^{(m)}(a)}{g^{(m)}(a)}. $$

Proof. Put $h=x-a$. Taylor’s Theorem, Peano Form gives

f(a+h)=\frac{f^{(m)}(a)}{m!}h^m+o(h^m), \qquad g(a+h)=\frac{g^{(m)}(a)}{m!}h^m+o(h^m).

For $h\ne0$, factor $h^m$ from each expression. The factors left after this cancellation tend respectively to $f^{(m)}(a)/m!$ and $g^{(m)}(a)/m!$. The latter limit is nonzero, so the denominator factor is nonzero whenever $h$ is sufficiently small. The quotient is therefore defined on a punctured neighborhood of $a$, and its limit is

\frac{f^{(m)}(a)/m!}{g^{(m)}(a)/m!} = \frac{f^{(m)}(a)}{g^{(m)}(a)}.

This proves both the nonvanishing claim and the limit formula. $\square$

Worked Example: A Logarithmic Limit After Three Cancellations

Evaluate

\lim_{x\to0} \frac{\ln(1+x)-x+x^2/2}{x^3}.

The logarithm is defined and has derivatives of every order near zero. Let $f(x)=\ln(1+x)-x+x^2/2$ and $g(x)=x^3$. Direct differentiation gives

f(0)=0,\qquad f'(x)=\frac{1}{1+x}-1+x,\qquad f'(0)=0,

f''(x)=-\frac{1}{(1+x)^2}+1,\qquad f''(0)=0, \qquad f'''(x)=\frac{2}{(1+x)^3},\qquad f'''(0)=2.

For the denominator, $g(0)=g'(0)=g''(0)=0$ and $g'''(0)=6$. The first nonzero order is three for both functions, so the quotient theorem yields

\lim_{x\to0}\frac{f(x)}{g(x)} = \frac{f'''(0)}{g'''(0)} = \frac{2}{6} = \frac13.

The cancellations in the numerator are reflected exactly in its first three Taylor coefficients: the constant, linear, and quadratic terms vanish. Comparing the cubic coefficients resolves the limit.

The equal-order hypothesis matters. When the numerator and denominator vanish to different orders, the quotient theorem above does not apply; one must compare the leading powers separately. Also, a quotient limit requires the denominator to be nonzero near the point on the punctured domain. The nonzero leading coefficient in the theorem guarantees this for the matching-order case.

Worked Example: A Square-Root Limit

Consider

\lim_{x\to0}\frac{\sqrt{1+x}-1-x/2}{x^2}.

Let $f(x)=\sqrt{1+x}-1-x/2$ and $g(x)=x^2$, both defined near zero. We have $f(0)=0$ and

f'(x)=\frac{1}{2\sqrt{1+x}}-\frac12,\qquad f'(0)=0, \qquad f''(x)=-\frac{1}{4(1+x)^{3/2}},\qquad f''(0)=-\frac14.

Meanwhile $g(0)=g'(0)=0$ and $g''(0)=2$. Both functions first have a nonzero derivative of order two at zero, so

\lim_{x\to0}\frac{\sqrt{1+x}-1-x/2}{x^2} = \frac{f''(0)}{g''(0)} = -\frac18.

Equivalently, the degree-two Taylor expansion says that the numerator’s leading term is $-x^2/8$. The negative sign records that the square root lies below its tangent-line approximation near zero.

A Remainder Estimate Can Prove an Inequality

The local sign theorem uses an expansion near its center, while Taylor’s Theorem with Lagrange remainder can give a sign on an entire interval. For a first-degree approximation, the remainder involves the second derivative at an intermediate point. If that derivative has a fixed sign throughout the segment, the direction of the approximation follows.

Worked Example: A Strict Logarithmic Inequality

We prove that

\ln(1+x)<x \qquad\text{for }-1<x,\quad x\ne0.

Apply Taylor’s Theorem of degree one to $f(t)=\ln(1+t)$ centered at $0$, with target $x$. The segment between $0$ and $x$ lies in $(-1,\infty)$. For every such nonzero $x$, the Lagrange form gives a point $c$ strictly between $0$ and $x$ such that

\ln(1+x)=f(0)+f'(0)x+\frac{f''(c)}{2}x^2.

Here $f(0)=0$, $f'(0)=1$, and $f''(t)=-1/(1+t)^2$. Since $c>-1$, we have $1+c>0$, so $f''(c)<0$. Also $x^2/2>0$. Consequently,

\ln(1+x) = x-\frac{x^2}{2(1+c)^2} < x.

The inequality is strict because the remainder is strictly negative for every $x\ne0$. At $x=0$, equality holds. The argument works on both sides of zero: the sign comes from $x^2$ and the second derivative, not from the sign of $x$.

Choosing the Right Taylor Application

These applications use Taylor information in two related but distinct ways. For a local sign or an extremum, the first nonzero term determines the behavior because the remaining error is smaller than that term. For a quotient limit, matching leading powers cancel, leaving the ratio of their coefficients. For an inequality over a larger interval, a Lagrange remainder can be more useful than a local expansion, because its intermediate point remains on the segment between the center and the target.

A common mistake is to stop at the first approximation without checking the error. A formal list of coefficients does not by itself establish a sign or a limit. Use the Peano remainder when the conclusion is local, and use an explicit remainder estimate when its sign or size is needed. This distinction is the basis for applying Taylor’s Theorem reliably.

Check Your Understanding

Use the leading-term principle and remainder estimates to check your understanding.

Why can a remainder $o(h^m)$ not change the sign of a nonzero leading term $ch^m$ for all sufficiently small nonzero $h$?
Under the local sign theorem, what does an even order $m$ imply when $f^{(m)}(a)<0$?
Why does the quotient theorem require the denominator’s first nonzero derivative to occur at the same order as the numerator’s?
For the square-root limit, which derivatives vanish at zero, and what is the ratio of the first nonzero derivatives?
In the logarithmic inequality, why is the Taylor remainder negative for targets on either side of zero?