Find lim as x to infinity (x^n)/(e^ax), where a>0 and n

Find $\displaystyle\lim_{x\to\infty}\frac{x^{n}}{e^{ax}}$, where $a>0$ and $n$ is a positive integer.

Step 1: The form

With $a>0$ both $x^{n}$ and $e^{ax}$ tend to $+\infty$, so the form is $\dfrac{\infty}{\infty}$.

Step 2: What each differentiation does

\[ \frac{d^{k}}{dx^{k}}x^{n} = \frac{n!}{\bigl(n-k\bigr)!}x^{\,n-k}, \qquad \frac{d^{k}}{dx^{k}}e^{ax} = a^{k}e^{ax}. \]

The numerator loses a power each time; the denominator only gains a constant factor $a$.

Step 3: Apply l'Hôpital's rule $n$ times

Every intermediate form is $\dfrac{\infty}{\infty}$, so the rule may be applied repeatedly until the numerator is constant: $\lim_{x\to\infty}\frac{x^{n}}{e^{ax}} = \lim_{x\to\infty}\frac{n!}{a^{n}e^{ax}}$.

Step 4: The last limit

\[ \frac{n!}{a^{n}}\lim_{x\to\infty}e^{-ax} = \frac{n!}{a^{n}}\cdot0 = 0, \]

using $a>0$.

Step 5: The condition $a>0$ is essential

\[a = 0:\ \frac{x^{n}}{1} = x^{n} \to +\infty\]

\[a\lt 0:\ e^{ax}\to0^{+} \text{ while } x^{n}\to+\infty, \text{ so the ratio} \to +\infty\]

\[n = 3,\ a = 2:\ \text{the limit is } 0; \text{ at } x = 10 \text{ the ratio is } \frac{1000}{4.85\cdot10^{8}} \approx 2\cdot10^{-6} \;\;\checkmark\]