We are asked to find the limit of the expression $\frac{n^{100}}{e^n}$ as $n$ approaches infinity.

1. Problem Description

We are asked to find the limit of the expression $\frac{n^{100}}{e^n}$ as $n$ approaches infinity.

2. Solution Steps

Since we are dealing with a limit of the form $\frac{\infty}{\infty}$, we can apply L'Hopital's Rule. L'Hopital's Rule states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the limit exists.

Let $f(n) = n^{100}$ and $g(n) = e^n$. Then $f'(n) = 100n^{99}$ and $g'(n) = e^n$.

We have $\lim_{n \to \infty} \frac{n^{100}}{e^n} = \lim_{n \to \infty} \frac{100n^{99}}{e^n}$.

We can apply L'Hopital's Rule repeatedly. After applying it 100 times, we get:

$\lim_{n \to \infty} \frac{n^{100}}{e^n} = \lim_{n \to \infty} \frac{100!}{e^n}$.

Since $100!$ is a constant and $e^n$ goes to infinity as $n$ goes to infinity, we have:

$\lim_{n \to \infty} \frac{100!}{e^n} = 0$.

Alternatively, using L'Hopital's rule $k$ times on $\lim_{n \to \infty} \frac{n^{100}}{e^n}$, where $k \le 100$

$\lim_{n \to \infty} \frac{n^{100}}{e^n} = \lim_{n \to \infty} \frac{100(99)(98) \dots (100-k+1) n^{100-k}}{e^n}$.

When $k=100$,

$\lim_{n \to \infty} \frac{n^{100}}{e^n} = \lim_{n \to \infty} \frac{100!}{e^n} = 0$.

3. Final Answer

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