Proving Limits With Taylor's Theorem

From Taylor Estimates to Limits

Taylor’s theorem gives more than a polynomial approximation: its remainder bound can show that an approximation error is small relative to the expression in a limit. This is especially useful when leading terms cancel. In that situation, simply substituting the limiting value may give $0/0$; the challenge is to expand far enough to identify the first term that survives.

Earlier in the course, Taylor’s Theorem, the Taylor Remainder Bound, and the Local Refinement of the Taylor Remainder were established. We will use those results rather than prove them again. The key practical decision is the order of expansion. If the denominator is of size $x^m$, an error bounded by a constant times $|x|^{m+1}$ becomes negligible after division by the denominator. If the expression has further cancellation, the expansion may need to go to a higher order.

Limit strategy: To evaluate a limit near $a$, write $h=x-a$, expand each relevant function at $a$, and keep every term that could survive the division or cancellation. Then use a Taylor remainder bound to justify that the omitted terms do not affect the limit.

A remainder estimate must match the scale of the limit. For example, an error bounded by $C|h|^2$ is not automatically negligible when the denominator is also of size $h^2$: after division, the estimate is only bounded by $C$. An error bounded by $C|h|^3$, on the other hand, becomes negligible after division by $h^2$.

When a Remainder Becomes Negligible

The following simple consequence of a Taylor remainder bound formalizes that comparison. The denominator may be any nonzero quantity, not necessarily a power of the displacement from the expansion point.

Theorem (Negligible Taylor Remainder Under Rescaling): Let $f$ have derivatives through order $n+1$ on an open interval containing $a$. Suppose $|f^{(n+1)}(t)|\leq M$ on a neighborhood of $a$, and let $$ P_n(h)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}h^k. $$ Suppose $u(x)\to0$, $v(x)\ne0$ near the limiting point, and $$ \frac{|u(x)|^{n+1}}{|v(x)|}\longrightarrow0. $$ Then $$ \frac{f(a+u(x))-P_n(u(x))}{v(x)}\longrightarrow0. $$

Proof. For $x$ sufficiently close to the limiting point, $a+u(x)$ lies in the neighborhood where the derivative bound holds. By the Taylor Remainder Bound,

|f(a+u(x))-P_n(u(x))| \leq \frac{M}{(n+1)!}|u(x)|^{n+1}.

Since $v(x)\ne0$, divide by $|v(x)|$ to obtain

\left|\frac{f(a+u(x))-P_n(u(x))}{v(x)}\right| \leq \frac{M}{(n+1)!}\frac{|u(x)|^{n+1}}{|v(x)|}.

The right-hand side tends to zero by hypothesis. The absolute value of the quotient is therefore bounded by a quantity tending to zero, so the quotient tends to zero. $\square$

This result is useful when the variable entering a Taylor expansion is itself a function of another variable, or when the denominator has a different scale from a simple power. In routine calculations, one often applies the Taylor Remainder Bound directly; the theorem clarifies exactly which comparison is needed.

Worked Example: The Limit of the Sinc Quotient

We prove that $\sin x/x\to1$ as $x\to0$. Expand $\sin x$ at $0$ through degree one. Since $\sin 0=0$ and $\cos 0=1$, the polynomial is $P_1(x)=x$. The second derivative is $-\sin t$, whose absolute value is at most $1$ for every real $t$. Taylor’s theorem therefore gives

|\sin x-x|\leq \frac{|x|^2}{2}.

For $x\ne0$, divide the error by $|x|$:

\left|\frac{\sin x}{x}-1\right| =\frac{|\sin x-x|}{|x|} \leq\frac{|x|}{2}.

The right-hand side tends to zero, so $\sin x/x\to1$. This proof shows why a first-order Taylor polynomial is enough: its error is of order $|x|^2$, which is negligible after division by $|x|$.

Expand Past the Cancellation

If several low-order terms cancel, the first-order estimate may be too coarse. The next example has a numerator in which both the constant and linear terms vanish. A second-order Taylor polynomial identifies the term that remains after division by $x^2$.

Worked Example: A Square-Root Remainder Limit

Consider

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

Let $f(x)=\sqrt{1+x}$, defined for $x>-1$. Its values and first two derivatives at $0$ are

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

Thus the degree-two Taylor polynomial at $0$ is $1+x/2-x^2/8$. To control its error, note that

f'''(x)=\frac{3}{8(1+x)^{5/2}}.

For $|x|\leq1/2$, we have $1+x\geq1/2$, so $|f'''(x)|\leq (3/8)2^{5/2}$. Taylor’s Remainder Bound gives, for some fixed constant $C$,

\left|\sqrt{1+x}-\left(1+\frac{x}{2}-\frac{x^2}{8}\right)\right| \leq C|x|^3.

For nonzero $x$, divide the numerator in the proposed limit by $x^2$, using the polynomial and its remainder:

\frac{\sqrt{1+x}-1-\frac{x}{2}}{x^2} =-\frac18+ \frac{\sqrt{1+x}-\left(1+\frac{x}{2}-\frac{x^2}{8}\right)}{x^2}.

The absolute value of the second term is at most $C|x|$, which tends to zero. The limit is therefore $-1/8$. The third derivative bound matters: it makes the error smaller than the $x^2$ scale of the denominator.

When combining different functions, it is often helpful to divide both numerator and denominator by the same power of $x$ before taking a limit. Taylor estimates can then be applied to each scaled expression, and the remaining denominator must be checked to have a nonzero limit.

Worked Example: A Quotient with Cancellation in Both Functions

Evaluate

\lim_{x\to0}\frac{e^x-1-x}{1-\cos x}.

For the numerator, expand $e^x$ at $0$ through degree two. Its third derivative is $e^t$, which is bounded by $e$ for $|t|\leq1$. Hence

e^x=1+x+\frac{x^2}{2}+R_1(x), \qquad |R_1(x)|\leq \frac{e}{6}|x|^3 \quad (|x|\leq1).

For the denominator, use the degree-three Taylor polynomial of $\cos x$ at $0$, which is $1-x^2/2$. The fourth derivative of cosine has absolute value at most $1$, so

\cos x=1-\frac{x^2}{2}+R_2(x), \qquad |R_2(x)|\leq \frac{|x|^4}{24}.

For $x\ne0$, the numerator divided by $x^2$ tends to $1/2$, because

\frac{e^x-1-x}{x^2}=\frac12+\frac{R_1(x)}{x^2}, \qquad \left|\frac{R_1(x)}{x^2}\right|\leq\frac{e}{6}|x|.

Likewise, the denominator divided by $x^2$ tends to $1/2$:

\frac{1-\cos x}{x^2}=\frac12-\frac{R_2(x)}{x^2}, \qquad \left|\frac{R_2(x)}{x^2}\right|\leq\frac{x^2}{24}.

In particular, $(1-\cos x)/x^2$ is nonzero for all sufficiently small nonzero $x$, since it tends to $1/2$. The original quotient is the ratio of these two scaled expressions, so its limit is $(1/2)/(1/2)=1$.

A Variable Power via the Logarithm

Taylor estimates also help with expressions whose exponent depends on the variable. The logarithm converts the power into a product, and a first-order Taylor estimate then identifies the limiting exponent. The domain restriction is important: for real powers defined through logarithms, the base must be positive.

Theorem (A Standard Variable-Power Limit): $$ \lim_{x\to0}(1+x)^{1/x}=e, $$ where the expression is considered for $x\ne0$ sufficiently close to $0$.

Proof. When $|x|<1/2$, the base $1+x$ is positive. Define

L(x)=\frac{\ln(1+x)}{x}.

For $f(x)=\ln(1+x)$, we have $f(0)=0$, $f'(0)=1$, and $f''(t)=-1/(1+t)^2$. On $[-1/2,1/2]$, $|f''(t)|\leq4$. Taylor’s Theorem through degree one therefore yields

|\ln(1+x)-x|\leq 2x^2.

After dividing by $|x|$, we obtain

|L(x)-1|=\left|\frac{\ln(1+x)-x}{x}\right|\leq2|x|.

Thus $L(x)\to1$. Since $1+x>0$, the definition of real powers gives

(1+x)^{1/x} =\exp\left(\frac{\ln(1+x)}{x}\right) =\exp(L(x)).

The exponential function is continuous, so $\exp(L(x))\to\exp(1)=e$. This proves the limit. $\square$

Choosing the Order and Avoiding Pitfalls

These examples suggest a reliable sequence of decisions. First identify which terms cancel. Then choose an expansion order that includes the first term that may remain after division. Finally, bound the remainder and compare it with the denominator. The expansion is not justified merely because a formal power series suggests the answer; the remainder estimate is what makes the argument rigorous.

Identify the scale.
Determine the power or other quantity by which the expression is divided, and note any terms that cancel.

Choose the Taylor order.
Keep enough terms to include the first contribution that can survive the cancellation and division.

Control the remainder.
Use a derivative bound on a neighborhood of the expansion point to estimate the error.

Finish the limit argument.
Show the scaled error tends to zero, and check that any denominator remaining in a quotient tends to a nonzero value.

A common mistake is to stop at an expansion whose remainder is only the same size as the denominator. For example, an estimate $|R(x)|\leq Cx^2$ does not show that $R(x)/x^2\to0$. Another is to divide two quantities that both tend to zero without first proving that the scaled denominator has a nonzero limit. The quotient example above addresses this by showing that the denominator divided by $x^2$ tends to $1/2$.

Taylor’s theorem is most effective in limit problems when used as an estimate rather than as a formal substitution rule. The polynomial reveals the candidate limit; the remainder bound verifies that higher-order terms cannot change it.

Check Your Understanding

Use the expansion order and remainder scale to decide how each argument should proceed.

Why is a degree-one Taylor polynomial sufficient to prove that $\sin x/x\to1$?
In the square-root example, why must the expansion include the quadratic term?
What remainder estimate is needed for an error to vanish after division by $x^2$?
Why is it necessary to show that $(1-\cos x)/x^2$ has a nonzero limit in the quotient example?
Why does taking the logarithm help evaluate $(1+x)^{1/x}$, and where is positivity of the base used?