Find lim as x to infinity x^ne^-x, where n in N
Find $\displaystyle\lim_{x\to\infty}x^{n}e^{-x}$, where $n\in\mathbb{N}$.
Explanation of Method
n differentiations exhaust the power; the exponential is untouched.
Step 1: Rewrite as a fraction
\[ x^{n}e^{-x} = \frac{x^{n}}{e^{x}}. \]
As $x\to\infty$ both $x^{n}\to+\infty$ and $e^{x}\to+\infty$: the form is $\dfrac{\infty}{\infty}$.
Step 2: Apply l'Hôpital's rule $n$ times
Each differentiation lowers the power in the numerator by one and leaves the denominator unchanged. After $k$ steps
\[ \frac{d^{k}}{dx^{k}}\bigl[x^{n}\bigr] = \frac{n!}{\bigl(n-k\bigr)!}\,x^{\,n-k}, \qquad \frac{d^{k}}{dx^{k}}\bigl[e^{x}\bigr] = e^{x}, \]
and every intermediate quotient is again of the form $\dfrac{\infty}{\infty}$, so the rule may be applied again. At $k = n$ the numerator has become the constant $n!$: $\lim_{x\to\infty}\frac{x^{n}}{e^{x}} = \lim_{x\to\infty}\frac{n!}{e^{x}}$.
Step 3: Evaluate the resulting limit
\[ \lim_{x\to\infty}\frac{n!}{e^{x}} = n!\cdot\lim_{x\to\infty}e^{-x} = n!\cdot0 = 0. \]
Step 4: Verification
\[n = 5:\quad x = 5:\ 21.056; \qquad x = 20:\ 6.596\cdot10^{-3}; \qquad x = 50:\ 6.03\cdot10^{-14} \;\;\checkmark\]
\[\text{The maximum of } x^{n}e^{-x} \text{ is at } x = n,\ \text{where it equals } n^{n}e^{-n} \;\;\checkmark\]
\[\text{Beyond } x = n \text{ the function decreases: } \bigl(x^{n}e^{-x}\bigr)' = x^{n-1}e^{-x}\bigl(n-x\bigr)\lt 0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\lim_{x\to\infty}x^{n}e^{-x} = 0, \qquad n\in\mathbb{N}\,}\]
Remark
Remark. The result holds for every real exponent, not only natural $n$: taking $m = \lceil n\rceil$ gives $x^{n}e^{-x}\le x^{m}e^{-x}\to0$ for large $x$. This is the precise sense in which the exponential outgrows every power — and by the substitution $x = \ln u$ it is the same statement as $\ln^{n}u/u\to0$, the logarithm losing to every power.