Let us apply the results of § 69 to the particularly important case in which \(\phi(n) = x^{n}\). If \(x = 1\) then \(\phi(n) = 1\), \(\lim\phi(n) = 1\), and if \(x = 0\) then \(\phi(n) = 0\), \(\lim \phi(n) = 0\), so that these special cases need not detain us.

First, suppose \(x\) positive. Then, since \(\phi(n + 1) = x\phi(n)\), \(\phi(n)\) increases with \(n\) if \(x > 1\), decreases as \(n\) increases if \(x < 1\).

If \(x > 1\), then \(x^{n}\) must tend either to a limit (which must obviously be greater than \(1\)) or to \(+\infty\). Suppose it tends to a limit \(l\). Then \(\lim\phi(n + 1) = \lim\phi(n) = l\), by Exs. XXV. 7; but \[\lim\phi(n + 1) = \lim x\phi(n) = x\lim\phi(n) = xl,\] and therefore \(l = xl\): and as \(x\) and \(l\) are both greater than \(1\), this is impossible. Hence \[x^{n} \to +\infty\quad (x > 1).\]

*Example.* The reader may give an alternative proof, showing by the binomial theorem that \(x^{n} > 1 + n\delta\) if \(\delta\) is positive and \(x = 1 + \delta\), and so that \[x^{n} \to +\infty.\]

On the other hand \(x^{n}\) is a decreasing function if \(x < 1\), and must therefore tend to a limit or to \(-\infty\). Since \(x^{n}\) is positive the second alternative may be ignored. Thus \(\lim x^{n} = l\), say, and as above \(l = xl\), so that \(l\) must be zero. Hence \[\lim x^{n} = 0\quad (0 < x < 1).\]

*Example.* Prove as in the preceding example that \((1/x)^{n}\) tends to \(+\infty\) if \(0 < x < 1\), and deduce that \(x^{n}\) tends to \(0\).

We have finally to consider the case in which \(x\) is negative. If \(-1 < x < 0\) and \(x = -y\), so that \(0 < y < 1\), then it follows from what precedes that \(\lim y^{n} = 0\) and therefore \(\lim x^{n} = 0\). If \(x = -1\) it is obvious that \(x^{n}\) oscillates, taking the values \(-1\), \(1\) alternatively. Finally if \(x < -1\), and \(x = -y\), so that \(y > 1\), then \(y^{n}\) tends to \(+\infty\), and therefore \(x^{n}\) takes values, both positive and negative, numerically greater than any assigned number. Hence \(x^{n}\) oscillates infinitely. To sum up: \[\begin{aligned} &\phi(n) = x^{n} \to +\infty &&(x > 1),\\ &\lim \phi(n) = 1 &&(x = 1),\\ &\lim \phi(n) = 0 &&(-1 < x < 1),\\ &\text{$\phi(n)$ oscillates finitely} &&(x = -1),\\ &\text{$\phi(n)$ oscillates infinitely}\qquad &&(x < -1).\end{aligned}\]

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