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 $$y$$.