18. Sequences

In general, the process of finding the limit of a sequence is essentially the same as finding the limit at infinity of a continuous function (i.e. finding its horizontal asymptote).

b3. Special Limits

Before discussing the computation of limits of sequences in general, we first present some Special Limits. The Special Limits are intuitively obvious, but the proofs may be found at the end of this chapter in the section on the Precise Definition of Limits.

The simplest two sequences are the identity sequence, \(a_n=n\), and the constant sequences, \(a_n=c\) for any number \(c\). Their plots are:

Their limits are given by:

\[ (1) \qquad \lim_{n\to\infty}n=\infty \]

For any number \(c\), \[ (2) \qquad \lim_{n\to\infty}c=c \]

The next two groups of sequences are the power sequences, \(a_n=n^p\), (with a constant power, \(p\)) and the exponential sequences, \(a_n=b^n\), (with a constant base, \(b\)).

If \(p>0\), then \[ (3a) \qquad \lim_{n\to\infty}n^p=\infty \qquad \text{and} \qquad (3b) \qquad \lim_{n\to\infty}\dfrac{1}{n^p}=0. \] If \(p<0\), then \[ (3c) \qquad \lim_{n\to\infty}n^p=0 \qquad \text{and} \qquad (3d) \qquad \lim_{n\to\infty}\dfrac{1}{n^p}=\infty. \]

As examples we have: \[\begin{array}{ll} \lim\limits_{n\to\infty}n^2 =\lim\limits_{n\to\infty}\dfrac{1}{n^{-2}}=\infty \qquad & \lim\limits_{n\to\infty}n^{1/2} =\lim\limits_{n\to\infty}\dfrac{1}{n^{-1/2}} =\lim\limits_{n\to\infty}\sqrt{n}=\infty \\ \lim\limits_{n\to\infty}n^{-2} =\lim\limits_{n\to\infty}\dfrac{1}{n^2}=0 \qquad & \lim\limits_{n\to\infty}n^{-1/2} =\lim\limits_{n\to\infty}\dfrac{1}{n^{1/2}} =\lim\limits_{n\to\infty}\dfrac{1}{\sqrt{n}}=0 \end{array}\] To justify these, try plugging in \(n=1\) million, i.e. \(n=1,000,000=10^6\).

If \(b>1\), then \[ (4a) \qquad \lim_{n\to\infty}b^n=\infty \qquad \text{and} \qquad (4b) \qquad \lim_{n\to\infty}\dfrac{1}{b^n}=0. \] If \(0 < b < 1\), then \[ (4c) \qquad \lim_{n\to\infty}b^n=0 \qquad \text{and} \qquad (4d) \qquad \lim_{n\to\infty}\dfrac{1}{b^n}=\infty. \]

As examples we have: \[\begin{array}{ll} \lim\limits_{n\to\infty}2^n=\infty \qquad &\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\right)^n =\lim\limits_{n\to\infty}\dfrac{1}{2^n}=0 \end{array}\] To justify these, try plugging in \(n=10\) or \(n=20\).

Study the difference between the power and the exponential sequences. The first is the limit of the variable base \(n\) to a fixed power \(p\). The second is the limit of a fixed base \(b\) to the variable power \(n\).