More on limits

We know that as x tends to $+\infty$, the greater the exponent in x^p, the faster x^p tends to infinity, in the sense that if 0 < q < p then

$$\lim_{x\rightarrow+\infty}\frac{x^q}{x^p} = 0$$

We now intend to show that

$$\lim_{x\rightarrow+\infty}\frac{x}{e^x} = 0$$

From this it will follow that for any positive number p,

$$\lim_{x\rightarrow+\infty}\frac{x^p}{e^x} = 0$$

since

$$\frac{x^p}{e^x} = \left(\frac{x}{e^{x/p}}\right)^p = p^p\left(\frac{x/p}{e^{x/p}}\right)^p$$

and x tends to infinity is the same as x/p tends to infinity since p > 0.

To show that

$$\lim_{x\rightarrow+\infty}\frac{x}{e^x} = 0$$

we will use the Funnel Theorem. We will show that for $x\geq 0$ that

$$1 + x + \frac{1}{2}x^2 \leq e^x.$$

This tells us that for x > 0 we have

$$0 \leq \frac{x}{e^x} \leq \frac{x}{(x^2/2) + x + 1}$$

Applying the Funnel Theorem to this inequality does the trick.

To get the inequality we need, first observe that the inequality is true when x = 0. Now consider what happens when x > 0. In this case, for $u\in[0,x]$ we have $1 \leq e^u$ so

$$\int_0^x 1 \;du \leq \int_0^x e^u\;du$$

Each of these integrals can be evaluated easily to get

$$x \leq e^x - e^0 = e^x - 1,$$

or

$$1 + x \leq e^x$$

for any positive real number x. Since this inequality also holds for x = 0 we have for any postive real number x, if $u\in[0,x]$ then

$$\int_0^x 1 + u \;du \leq \int_0^x e^u\;du$$

Employing the Fundamental Theorem of Calculus, we can evaluate these integrals to get

$$x + \frac{1}{2}x^2 - (0 + \frac{1}{2}0^2) \leq e^x - e^0$$

which is the same as

$$1 + x + \frac{1}{2}x^2 \leq e^x$$

just as we wished. Note that this process can be repeated to get for $x\geq 0$

$$\begin{array} {rcl} {\displaystyle 1 + x + \frac{1}{2}x^2 + ... ...rac{1}{6}x^3 + \cdots + \frac{1}{N!}x^N}& \leq & e^x\end{array}$$