The following limit shows that the growth of the exponential function \( e^x \) is much faster than any polynomial \( x^n \). This result is fundamental in mathematical analysis and is used in the theory of orders of magnitude and computational complexity.
Let's prove that:
\[ \lim_{x \to \infty} \frac{e^x}{x^n} = \infty \]
L'Hôpital's Rule
Let's consider the function:
\[ f(x) = \frac{e^x}{x^n}. \]
We apply L'Hôpital's rule, differentiating the numerator and denominator \( n \) times:
- The numerator \( e^x \) remains unchanged.
- The denominator, after \( n \) differentiations, becomes \( n! \).
Therefore, we have:
\[ \lim_{x \to \infty} \frac{e^x}{x^n} = \lim_{x \to \infty} \frac{e^x}{n!} = \infty. \]
Comparison of Orders of Magnitude
Another way to prove the limit is to note that the function \( e^x \) can be written as its Taylor series:
\[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}. \]
Let's consider only the term with \( k = n+1 \):
\[ e^x > \frac{x^{n+1}}{(n+1)!}. \]
Therefore:
\[ \frac{e^x}{x^n} > \frac{x^{n+1}}{(n+1)! x^n} = \frac{x}{(n+1)!}. \]
Since \( \displaystyle \frac{x}{(n+1)!} \to \infty \) as \( x \to \infty \), it follows that \( \displaystyle \frac{e^x}{x^n} \to \infty \).
Asymptotic Method
We can also use an asymptotic approach. We observe that the ratio between the two functions is:
\[ \frac{e^x}{x^n} = e^{x - n \ln x}. \]
For sufficiently large \( x \), the term \( x - n \ln x \) grows indefinitely, therefore \( e^{x - n \ln x} \to \infty \), and the limit is proven.