\[e^x\]
tends to infinity as \[x\]
tends to infinity faster than any power of \[x\]
.We can show this using l'Hospital's Rule repetedly, since
\[lim_{x \rightarrow \infty} \frac{e^x}{x^n}\]
is of indeterminate form.\[\begin{equation} \begin{aligned} lim_{x \rightarrow \infty} \frac{e^x}{x^n} &= lim_{x \rightarrow \infty}\frac{e^x}{nx^{n-1}} \\ &= lim_{x \rightarrow \infty}\frac{e^x}{n(n-1)x^{n-2}} \\ &= \vdots \; \; \; \; \vdots \; \; \; \; \vdots \; \; \; \; \\ &= lim_{x \rightarrow \infty}\frac{e^x}{n!x^0} \\ &= \infty \end{aligned} \end{equation}\]
.We can also show this with Mclaurin series.
\[\begin{equation} \begin{aligned} lim_{x \rightarrow \infty} \frac{e^x}{x^n} &= \frac{ \sum_{k=0}^{\infty} \frac{x^k}{k!}}{x^n} \\ &= \sum_{k=0}^{\infty} \frac{x^{k-n}}{k!} \end{aligned} \end{equation}\]
There are an infinite number of terms with positive power, each of which tends to infinity as
\[x\]
tends to infinity. Each term of the summation is positive for positive \[x\]
so the sum and \[lim{x \rightarrow \infty} \frac{e^x}{x^n}\]
tend to infity and the statement is true.