We can also prove (cf. § 75) that \[\lim n(1 – x^{-1/n}) = \lim n(x^{1/n} – 1) = \log x.\]

For \[n(x^{1/n} – 1) – n(1 – x^{-1/n}) = n(x^{1/n} – 1)(1 – x^{-1/n}),\] which tends to zero as \(n \to \infty\), since \(n(x^{1/n} – 1)\) tends to a limit (§ 75) and \(x^{-1/n}\) to \(1\) (Ex. XXVII. 10). The result now follows from the inequalities (3) of § 208.

Example LXXXVI
1. Prove, by taking \(y = 1\) and \(n = 6\) in the inequalities (4) of § 208, that \(2.5 < e < 2.9\).

2. Prove that if \(t > 1\) then \((t^{1/n} – t^{-1/n})/(t – t^{-1}) < 1/n\), and so that if \(x > 1\) then \[\int_{1}^{x} \frac{dt}{t^{1-(1/n)}} – \int_{1}^{x} \frac{dt}{t^{1+(1/n)}} < \frac{1}{n} \int_{1}^{x} \left(t – \frac{1}{t}\right) \frac{dt}{t} = \frac{1}{n} \left(x + \frac{1}{x} – 2\right).\] Hence deduce the results of § 209.

3. If \(\xi_{n}\) is a function of \(n\) such that \(n\xi_{n} \to l\) as \(n \to \infty\), then \((1 + \xi_{n})^{n} \to e^{l}\). [Writing \(n\log(1 + \xi_{n})\) in the form \[l \left(\frac{n\xi_{n}}{l}\right) \frac{\log(1 + \xi_{n})}{\xi_{n}},\] and using Ex. LXXXII. 4, we see that \(n\log(1 + \xi_{n})\to l\).]

4. If \(n\xi_{n} \to \infty\), then \((1 + \xi_{n})^{n} \to \infty\); and if \(1 + \xi_{n} > 0\) and \(n\xi_{n} \to -\infty\), then \[(1 + \xi_{n})^{n} \to 0.\]

5. Deduce from (1) of § 208 the theorem that \(e^{y}\) tends to infinity more rapidly than any power of \(y\).


$\leftarrow$ 208. The representation of \(e^{x}\) as a limit Main Page 210. Common logarithms $\rightarrow$
Close Menu