The Approximation Strategy

Prove the Claim Through a Simpler Object

The Cauchy Strategy established convergence by controlling distances between late terms, without requiring the limit to be known in advance. The Approximation Strategy takes a different route: replace an object by a simpler one, estimate the error, and use that estimate to transfer a property from the simpler object to the original one.

The strategy is useful when the original expression is awkward but a nearby expression is familiar. A proof might compare a sequence to a known convergent sequence, or compare a function to a continuous function. The key is not merely to say that the approximation is “close.” The proof must specify how close it is, and the error bound must be strong enough to establish the desired conclusion.

We begin with a basic stability result for sequences. It explains precisely why a perturbation that tends to zero cannot change a limit. We then consider a function version in which one continuous function approximates another uniformly. Each result turns an approximation estimate into a rigorous transfer principle.

Vanishing Errors Preserve Sequence Limits

Theorem (Stability of Sequence Limits Under Vanishing Perturbations): Let \((x_n)\) and \((y_n)\) be real sequences. Suppose \(y_n\to L\) and \(|x_n-y_n|\to0\). Then \(x_n\to L\).

Proof. Let \(\varepsilon>0\). Since \(y_n\to L\), there is an integer \(N_1\) such that \(n\geq N_1\) implies \(|y_n-L|<\varepsilon/2\). Since \(|x_n-y_n|\to0\), there is an integer \(N_2\) such that \(n\geq N_2\) implies \(|x_n-y_n|<\varepsilon/2\). Choose \(N=\max\{N_1,N_2\}\). For every \(n\geq N\), the triangle inequality gives

|x_n-L| \leq |x_n-y_n|+|y_n-L| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

Thus \(x_n\to L\). \(\square\)

This argument has a useful proof pattern. Split the allowed error into two parts: one part pays for the approximation, and the other pays for the simpler sequence’s distance to the limit. The two thresholds may be different, but taking their maximum produces a single index that works for both estimates.

Worked Example: A Small Oscillation Around a Familiar Sequence

Define \(x_n=1/n+(-1)^n/n^2\) and \(y_n=1/n\) for positive integers \(n\). The sequence \(y_n\) converges to \(0\), and the approximation error satisfies

|x_n-y_n| =\left|\frac{(-1)^n}{n^2}\right| =\frac{1}{n^2}.

Given \(\varepsilon>0\), choose \(N>1/\sqrt{\varepsilon}\). For \(n\geq N\), \(1/n^2\leq1/N^2<\varepsilon\), so \(|x_n-y_n|\to0\). The stability theorem now gives \(x_n\to0\). The alternating term need not be analyzed separately: its entire effect is bounded by an error that vanishes.

Worked Example: Estimating a Difference of Square Roots

Consider \(x_n=\sqrt{n^2+3n}-n\). The expression includes two large terms whose difference is not immediately transparent. Rationalizing gives an exact simpler form:

x_n =\frac{(\sqrt{n^2+3n}-n)(\sqrt{n^2+3n}+n)} {\sqrt{n^2+3n}+n} =\frac{3n}{\sqrt{n^2+3n}+n} =\frac{3}{\sqrt{1+3/n}+1}.

Use \(y_n=3/2\) as a comparison sequence. For \(t\geq0\), \(\sqrt{1+t}\geq1\), and \(\sqrt{1+t}-1=t/(\sqrt{1+t}+1)\leq t/2\). Therefore

\left|\frac{3}{\sqrt{1+t}+1}-\frac{3}{2}\right| =\frac{3}{2}\frac{\sqrt{1+t}-1}{\sqrt{1+t}+1} \leq\frac{3t}{8}.

Substitute \(t=3/n\). The algebraic identity and bound give

|x_n-y_n| =\left|\frac{3}{\sqrt{1+3/n}+1}-\frac{3}{2}\right| \leq\frac{9}{8n}.

Since \(9/(8n)\to0\) and \(y_n=3/2\) converges to \(3/2\), the stability theorem yields \(x_n\to3/2\). The estimate is what justifies replacing the original sequence by its simpler comparison sequence.

Uniform Approximation Transfers Continuity

For functions, an approximation estimate must often hold at every point in the domain at once. This is the role of uniform approximation. It lets one choose a single approximating function whose error is small throughout the domain, including at the nearby points used in a continuity proof.

Definition (Uniform Convergence): Let \(E\) be a set, and let \(f_n:E\to\mathbb{R}\) and \(f:E\to\mathbb{R}\). The sequence \((f_n)\) converges uniformly to \(f\) on \(E\) if, for every \(\varepsilon>0\), there is an integer \(N\) such that \(|f_n(x)-f(x)|<\varepsilon\) for every \(n\geq N\) and every \(x\in E\).

The decisive phrase is “every \(x\in E\).” The index \(N\) may depend on \(\varepsilon\), but it cannot depend on the point \(x\). This uniform control gives the following theorem.

Theorem (Uniform Limit of Continuous Functions Is Continuous): Let \(E\) be a subset of \(\mathbb{R}\), and let \(f_n:E\to\mathbb{R}\) be continuous at \(a\in E\) for every \(n\). If \(f_n\) converges uniformly on \(E\) to \(f:E\to\mathbb{R}\), then \(f\) is continuous at \(a\). In particular, if every \(f_n\) is continuous on \(E\), then \(f\) is continuous on \(E\).

Proof. Let \(\varepsilon>0\). By uniform convergence, choose \(n\) such that \(|f_n(x)-f(x)|<\varepsilon/3\) for every \(x\in E\). This single \(n\) works both at \(a\) and at all other points of \(E\). Since \(f_n\) is continuous at \(a\), there is a \(\delta>0\) such that, for \(x\in E\) with \(|x-a|<\delta\),

|f_n(x)-f_n(a)|<\frac{\varepsilon}{3}.

For any such \(x\), insert the approximating function’s values between \(f(x)\) and \(f(a)\). The triangle inequality gives

|f(x)-f(a)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)| <\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} =\varepsilon.

This is the definition of continuity of \(f\) at \(a\). Since \(a\) was any point where all the \(f_n\) are continuous, the final assertion follows. \(\square\)

Worked Example: Uniformly Approximating a Polynomial

On \(E=[-2,2]\), let \(f_n(x)=x^2+x/n\) and \(f(x)=x^2\). Every \(f_n\) is continuous, and for every \(x\in[-2,2]\),

|f_n(x)-f(x)| =\left|\frac{x}{n}\right| \leq\frac{2}{n}.

Given \(\varepsilon>0\), choose \(N>2/\varepsilon\). For \(n\geq N\) and every \(x\in[-2,2]\), the error is at most \(2/n\leq2/N<\varepsilon\). Thus \(f_n\) converges uniformly to \(f\) on this interval. The theorem guarantees continuity of the limit; directly, \(f(x)=x^2\) is a polynomial. The uniform estimate is stronger than checking convergence separately at each fixed \(x\), because it controls all points with the same \(N\).

Why Pointwise Closeness Is Not Enough

A common pitfall is to use an approximation that is good at each fixed point but not uniformly good across the domain. Pointwise convergence means that for each \(x\), the threshold \(N\) may depend on \(x\). That dependence can prevent the continuity proof from choosing one approximating function that works both at \(a\) and at nearby points.

Worked Example: Pointwise Convergence Without Uniform Convergence

On \(E=[0,1]\), define \(f_n(x)=x^n\). For each fixed \(x\) with \(0\leq x<1\), \(x^n\to0\), while \(f_n(1)=1\) for every \(n\). The pointwise limit is therefore

f(x)= \begin{cases} 0,&0\leq x<1,\\ 1,&x=1. \end{cases}

Each \(f_n\) is continuous on \([0,1]\), but \(f\) is not continuous at \(1\): for example, \(x_k=1-1/k\) tends to \(1\) as \(k\to\infty\), while \(f(x_k)=0\) for \(k\geq2\), not \(1=f(1)\).

The convergence cannot be uniform. For every \(n\), choose \(x_n=1-1/(n+1)\). Then \(0\leq x_n<1\), so \(f(x_n)=0\), and

|f_n(x_n)-f(x_n)| =\left(1-\frac{1}{n+1}\right)^n =\left(\frac{n}{n+1}\right)^n \geq\frac{1}{3}.

For the last inequality, the binomial theorem gives \((1+1/n)^n\leq\sum_{k=0}^{n}1/k!<3\); hence \((n/(n+1))^n=(1+1/n)^{-n}>1/3\). Thus the error is not uniformly small: no matter how large \(n\) is, some point of the interval has error greater than \(1/3\). This example shows why the uniform hypothesis in the continuity theorem cannot simply be replaced by pointwise convergence.

A Practical Approximation Checklist

An approximation proof should make three choices explicit: the simpler object, the error to be bounded, and the property that will transfer. For a sequence limit, the error should tend to zero as the index increases. For the continuity theorem, the approximation error must be uniformly small over the domain. Once the estimate is written, the triangle inequality often supplies the final step.

Choose the comparison object.
Select a sequence or function whose relevant behavior is already known or easier to prove.

Write an explicit error estimate.
Bound the distance between the original object and its approximation, keeping track of which variables the bound depends on.

Check the required strength.
For a sequence limit, the error must vanish; for a uniform-limit continuity argument, one threshold must work at every point.

Transfer the property.
Use the triangle inequality to combine the approximation error with the estimate for the simpler object.

Approximation is not a substitute for proof; it is a way to organize one. A comparison is useful only when the error estimate has the right form and the relevant hypotheses have been checked. Vanishing perturbations preserve sequence limits, while uniform approximation preserves continuity of limits. Keeping those two forms of control distinct is the central discipline of this strategy.

Check Your Understanding

Use the estimates and arguments in this tutorial to answer these questions.

In the sequence stability theorem, why are two separate error bounds needed before applying the triangle inequality?
What does uniform convergence require that pointwise convergence does not?
Where does the proof that a uniform limit is continuous use the fact that the approximation is uniform?
Why does the sequence \(x_n=1-1/(n+1)\) show that \(f_n(x)=x^n\) does not converge uniformly to its pointwise limit?
For what kinds of claims would an error bound depending on the point \(x\) be insufficient?

What You'll Learn