Call Us 07766496223
\[e^x\]
  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.