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:
\[ \lim_{x \to \infty} \frac{x^n}{e^x} = 0 \]
L'Hôpital's Rule
Let's consider the function:
\[ f(x) = \frac{x^n}{e^x} \]
We apply L'Hôpital's rule, differentiating the numerator and denominator \( n \) times:
- The numerator, after \( n \) differentiations, becomes \( n! \);
- The denominator \( e^x \) remains unchanged.
Therefore, we have:
\[ \lim_{x \to \infty} \frac{x^n}{e^x} = \lim_{x \to \infty} \frac{n!}{e^x} = 0 \]
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 the terms up to \( k = 2n \):
\[ e^x > 1 + x + \frac{x^2}{2!} + ... + \frac{x^{2n}}{(2n)!} > \frac{x^{2n}}{(2n)!} \]
Therefore:
\[ \frac{x^n}{e^x} < \displaystyle \frac{x^n}{\displaystyle \frac{x^{2n}}{(2n)!}} = \frac{(2n)!}{x^n} \]
Since \( \displaystyle \frac{(2n)!}{x^n} \to 0 \) as \( x \to \infty \), it follows that \( \displaystyle \frac{x^n}{e^x} \to 0 \).
Asymptotic Method
We can also use an asymptotic approach. We observe that the ratio between the two functions is:
\[ \frac{x^n}{e^x} = e^{n \ln x - x} \]
For sufficiently large \( x \), the term \( n \ln x - x \) becomes negative and decreases indefinitely, therefore \( e^{n \ln x - x} \to 0 \), and the limit is proven.