proof of growth of exponential function

In this proof, we first restrict to when $x$ and $a$ are integersand only later lift this restricton.

Let $a>0$ be an integer, let $b>1$ be real, and let $x$ be aninteger.

Consider the following inequality

$${\left(1+\frac{1}{x}\right)}^a\le 1+\frac{a}{x}{\left(1+\frac{1}{x}\right)}^{a-1}$$

If $x\ge 2$, then we have

$${\left(1+\frac{1}{x}\right)}^a\le 1+\frac{a}{x}{\left(\frac{3}{2}\right)}^{a-1}.$$

Define $X$ to be the greater of $2$ and $\lceil a{(3/2)}^{a-1}/(1-\sqrt{b})\rceil$; when $x>X$, we have

$${\left(1+\frac{1}{x}\right)}^a\le \sqrt{b}.$$

Rewrite $x^a/b^x$ as follows when $x>X$:

$$\frac{x^a}{b^x}=\frac{X^a}{b^X}\prod _{n=X}^x{\left(1+\frac{1}{n}\right)}^a\frac{1}{b}$$

By the inequality established above, each term in the product will bebounded by $1/\sqrt{b}$, hence

$$\frac{x^a}{b^x}\le \frac{X^a}{b^X}\frac{1}{{(\sqrt{b})}^{x-X}}$$

Since $b>1$, it is also the case that $\sqrt{b}>1$, hence we havethe inequality

$${(\sqrt{b})}^n\ge 1+n(\sqrt{b}-1)$$

Combining the last two inequalities yields the following:

$$\frac{x^a}{b^x}\le \frac{X^a}{b^X}\le \frac{1}{1+(x-X)(\sqrt{b}-1)}$$

From this, it follows that ${lim}_{x\to \mathrm{\infty }}{x}^a/b^x=0$ when$a$ and $x$ are integers.

Now we lift the restriction that $a$ be an integer. Since the powerfunction is increasing, $x^a/b^x\le x^{\lceil a\rceil }/b^x$, so we have ${lim}_{x\to \mathrm{\infty }}{x}^a/b^x=0$ for real valuesof $a$ as well.

To lift the restriction on $x$, let us write $x=x_1+x_2$ where$x_1$ is an integer and . Then we have

$$\frac{x^a}{b^x}=\frac{x_1^a}{b^{x_1}}{\left(\frac{x_1+x_2}{x_1}\right)}^a{b}^{-x_2}$$

If $x>2$, then . Since $x_2\ge 0,b^{-x_2}\le 1$. Hence, for all real $x>2$, we have

$$\frac{x^a}{b^x}\le {1.5}^a\frac{x_1^a}{b^{x_1}}$$

From this inequality, itfollows that ${lim}_{x\to \mathrm{\infty }}{x}^a/b^x=0$ for real values of$x$ as well.