Limit Properties

As before, “sequences” here implicitly mean “in an ordered field”. Where a complete ordered field is assume, it will be explicitly noted.

Assume that \(a_n\) and \(b_n\) are sequences which converge to \(a\) and \(b\) respectively.

\[-a_n-(-a)=-a_n+a=a-a_n=-(a_n-a)\]

and so \(|a_n-a|=|(-a_n)-(-a)|\), so \(-a_n\) converges to \(a\).

\[|(a_n+b_n)-(a+b)|=|(a_n-a)+(b_n-b)|\leq |a_n-a|+|b_n-b|\]

If we choose \(N\) such that the appropriate tails of \(a_n\) and \(b_n\) are :math:epsilon/2:-close to \(a\) and \(b\) respectively, then the sum above will be \(<\epsilon/2+\epsilon/2=\epsilon\). Therefore, \(a_n+b_n\) converges to \(a+b\) (or in other words, the limit of a sum is the sum of the limits). This technique is known as the “half-\(\epsilon\), although in many applications, dividing by more than two is necessary.

What about multiplication?

\[|a_nb_n-ab|=|a_nb_n-a_nb+a_nb-ab|\leq|a_n(b_n-b)|+|(a_n-a)b|=|a_n||b_n-b|+|b||a_n-a|\]

Note that since \(a_n\) converges, it is bounded from both above and below, and so \(|a_n|\) is bounded by a number, \(M\).

Define \(N\) such that the \(N\)-tail has \(|b_n-b|<\epsilon/2(M+1)\) and \(|a_n-a|`<\epsilon/2(|b|+1)\) (we add \(1\) to avoid dividing by zero). Then for \(n>N\)

\[|a_n||b_n-b|+|b||a_n-a| \leq M|b_n-b|+|b||a_n-a| < M\epsilon/2(M+1)+|b|\epsilon/2(|b|+1) < \epsilon\]

So the limit of a product is the product of the limits.

If \(a_n\) converges to \(a>0\), then a tail will be bounded from below by \(a-a/2=a/2\). Since the following only deals with tail properties, a simplifying assumption is that \(a_n>a/2\) for all \(n\). In particular, \(a_n\ne 0\) and so \(a_n^{-1}\) is well defined.

\[|a_n^{-1}-a^{-1}|=|a-a_n|/(|a_n|a|)<|a-a_n|/(2|a|^2)\]

If we take \(N\) for \(a_n\) to be \(\epsilon(2|a|^2)\)-close to \(a\), then the expression above is \(<\epsilon\).

If \(a<0\), then \(-a_n\) converges to \(-a>0\), \(1/-a_n=-1/a_n\) converges to \(1/-a=-1/a' and so :math:`1/a_n\) converges to \(1/a\). In summary, the limit of the inverse is the inverse of the limit.

If \(b>a\), let \(\epsilon=(b-a)/2\) and find \(N\) where for \(n>N\) both \(a_n\) is \(\epsilon\)-close to \(a\) and \(b_n\) is \(\epsilon\)-close to \(b\), then \(b_n-a_n>b-\epsilon-(a+\epsilon)=b-a-2\epsilon=d-d=0\), so \(b_n>a_n\) at the tail. In particular, if \(a_n\geq b_n\) on a tail, then it cannot be the case that \(b>a\) so \(a\geq b\).

The summary of these results is that in an ordered field, limits preserve the ordered field structure (weakly, for the order).

Assume \(a_n\) and \(c_n\) both converge to \(a\), and \(a_n\leq b_n\leq c_n\). For a tail, \(a-\epsilon<a_n\leq b_n \leq c_n<a+\epsilon\), so \(b_n\) also converges to \(a\). This result is called “the sandwich theorem”, and is often a powerful aid to finding and proving limits.