Limits Involving Absolute Values

When Can an Absolute Value Be Removed?

Absolute values can make a sequence look more complicated than it is. If the expression inside the bars is eventually positive, the bars do nothing from that point onward; if it is eventually negative, the bars simply change its sign. The key question is therefore whether the expression has a stable sign near its limit.

The previous tutorial proved the Squeeze Theorem, which is useful when direct estimates are needed. Here we use the epsilon–N definition and earlier limit laws to develop a complementary technique: a sequence converging to a nonzero number must eventually have the same sign as that number. This lets us simplify many absolute-value limits before evaluating them.

Proof. Let \(\varepsilon=|A|/2\), which is positive because \(A\neq0\). Since \(x_n\to A\), there is an index \(N\) such that \(|x_n-A|<|A|/2\) whenever \(n\geq N\).

If \(A>0\), then \(|A|=A\), and the inequality \(x_n>A-|A|/2=A/2>0\) holds for every \(n\geq N\). If \(A<0\), then \(|A|=-A\), and \(x_n<A+|A|/2=A/2<0\) for every \(n\geq N\). Thus in either case \(x_n\) eventually has the same strict sign as \(A\). \(\square\)

The nonzero hypothesis matters. A sequence converging to zero need not settle on one side of zero: its terms may be positive, negative, or zero in any pattern. Near a nonzero limit, however, the sequence eventually remains in an interval that does not contain zero.

Limits of Absolute-Value Expressions

Suppose \(a_n\to L\), and consider \(|a_n-c|\) for a fixed real number \(c\). By the Limit of a Difference Theorem, \(a_n-c\to L-c\), since the constant sequence with value \(c\) converges to \(c\). If \(L\neq c\), the limit \(L-c\) is nonzero. Eventual Sign Stability then tells us that \(a_n-c\) eventually has the sign of \(L-c\). Consequently, the absolute value can eventually be written without bars.

Proof. The Limit of a Difference Theorem gives \(a_n-c\to L-c\). If \(L>c\), then \(L-c>0\), so Eventual Sign Stability gives an index \(N\) such that \(a_n-c>0\) whenever \(n\geq N\). The definition of absolute value then gives \(|a_n-c|=a_n-c\) for those indices. Since \(a_n-c\to L-c=|L-c|\), and sequences that agree on a tail have the same limit, \(|a_n-c|\to|L-c|\).

If \(L<c\), then \(L-c<0\). Eventual Sign Stability gives \(a_n-c<0\) for all sufficiently large \(n\), so \(|a_n-c|=-(a_n-c)=c-a_n\) on that tail. The Limit of a Difference Theorem gives \(c-a_n\to c-L=|L-c|\), proving the conclusion in this case as well. \(\square\)

The limit conclusion also follows from the earlier Theorem (Absolute Values Preserve Limits), applied to \(a_n-c\). The additional information here is the eventual equality: once the sign is known, the absolute-value expression is eventually an ordinary difference. This can make subsequent algebra and estimates simpler.

Absolute Errors Can Converge Even When Their Signs Change

An absolute-value expression can also describe the size of an error. If \(a_n\to L\), then the error from \(L\) tends to zero, and its absolute value measures the distance from \(a_n\) to \(L\). The sign of the error may alternate without affecting that distance.

This example is an instance of the earlier Absolute-Value Criterion for Convergence to Zero: a sequence converges to zero exactly when its sequence of absolute values converges to zero. That criterion is special to the limit zero. For a nonzero magnitude limit, more information about the signs may be needed.

When Convergence of the Magnitudes Is Enough

Suppose \(|a_n|\to A\). If \(a_n\) is eventually nonnegative, then \(a_n=|a_n|\) on a tail, so \(a_n\to A\). If \(a_n\) is eventually nonpositive, then \(a_n=-|a_n|\) on a tail, so \(a_n\to-A\). This gives a useful converse-style result when sign information is available.

Proof. In the first case, choose \(N\) such that \(a_n\geq0\) for every \(n\geq N\). Then \(a_n=|a_n|\) whenever \(n\geq N\). Given \(\varepsilon>0\), convergence of \(|a_n|\) to \(A\) supplies an index \(N_1\) such that \(\bigl||a_n|-A\bigr|<\varepsilon\) whenever \(n\geq N_1\). For every \(n\geq\max\{N,N_1\}\), the tail equality gives

This is the epsilon–N condition for \(a_n\to A\).

In the second case, choose \(N\) such that \(a_n\leq0\) for every \(n\geq N\). Then \(a_n=-|a_n|\) whenever \(n\geq N\). For \(n\geq\max\{N,N_1\}\), where \(N_1\) is chosen as above, we have

Therefore \(a_n\to-A\). Both conclusions follow directly from the definition of convergence. \(\square\)

Why Magnitudes Alone May Not Determine a Limit

A limit for \(|a_n|\) does not generally determine a limit for \(a_n\). Absolute value discards the sign, and that lost information can matter. For instance, let

Since \(2+1/(n+2)>0\), its absolute value is

Nevertheless, \(a_n\) does not converge. Its even-indexed terms are \(2+1/(n+2)\) and tend to \(2\), while its odd-indexed terms are \(-2-1/(n+2)\) and tend to \(-2\). If the full sequence converged, every subsequence would converge to the same limit, by the Theorem (Subsequences of a Convergent Sequence). The even and odd subsequences have distinct limits, so the full sequence cannot converge.

There is no contradiction with the result about convergence to zero. If \(|a_n|\to0\), then the Absolute-Value Criterion for Convergence to Zero guarantees \(a_n\to0\) without any sign condition. But when \(|a_n|\to A\) for \(A>0\), the sequence may alternate between values near \(A\) and values near \(-A\). Eventual sign information is what distinguishes those possibilities.

A reliable approach is to identify the limit of the expression inside the bars. If that limit is nonzero, Eventual Sign Stability tells you which formula applies eventually. If the inside expression tends to zero, the absolute-value criterion or a direct error estimate is appropriate. If you know only that the magnitudes tend to a positive number, check whether the signs eventually settle before claiming that the original sequence converges.