Lemma. $$\lim_{x\to\infty}\frac{x^a}{e^x} = 0$$ for all constant values of $a$ .

Proof. Let $\varepsilon$ be any positive number. Then we get:

$$0 < \frac{x^a}{e^x} \leqq \frac{x^{\lceil a \rceil}}{e^x} < \frac{x^{\lceil a \rceil}}{\frac{x^{\lceil a\rceil+1}}{(\lceil a\rceil+1)!}} = \frac{(\lceil a\rceil+1)!}{x} < \varepsilon$$ as soon as $x > \max\{1, \frac{(\lceil a\rceil+1)!}{\varepsilon}\}$ . Here, $\lceil\cdot\rceil$ means the ceiling function; $e^x$ has been estimated downwards by taking only one of the all positive terms of the series expansion $$e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots$$

Proof. Since $\ln b > 0$ , we obtain by using the lemma the result $$\lim_{x\to\infty}\frac{x^a}{b^x} = \lim_{x\to\infty}\left(\frac{x^{\frac{a}{\ln b}}}{e^x}\right)^{\ln b} = 0^{\ln b} = 0.$$

Corollary 1. $\displaystyle\lim_{x\to 0+}x\ln{x} = 0.$

Proof. According to the lemma we get $$0 = \lim_{u\to\infty}\frac{-u}{e^u} = \lim_{x\to 0+}\frac{-\ln{\frac{1}{x}}}{\frac{1}{x}} = \lim_{x\to 0+}x\ln{x}.$$

Corollary 2. $\displaystyle\lim_{x\to\infty}\frac{\ln{x}}{x} = 0.$

Proof. Change in the lemma $x$ to $\ln{x}$ .

Corollary 3. $\displaystyle\lim_{x\to\infty}x^{\frac{1}{x}} = 1.$ (Cf. limit of nth root of n.)

Proof. By corollary 2, we can write: $\displaystyle x^{\frac{1}{x}} = e^{\frac{\ln{x}}{x}}\longrightarrow e^0 = 1$ as $x\to\infty$ (see also theorem 2 in limit rules of functions).