# What is the limit of (x-1)(x-2)(x-3)(x-4)(x-5)/(5x-1)^5 as x goes to infinity?

Sep 27, 2015

You can separate this into multiple limits using their multiplicative properties.

$= {\lim}_{x \to \infty} \frac{x - 1}{5 x - 1} \cdot {\lim}_{x \to \infty} \frac{x - 2}{5 x - 1} \cdot {\lim}_{x \to \infty} \frac{x - 3}{5 x - 1} \cdot {\lim}_{x \to \infty} \frac{x - 4}{5 x - 1} \cdot {\lim}_{x \to \infty} \frac{x - 5}{5 x - 1}$

Then, we can "simplify" this down using the fact that as $x$ becomes really large, the evaluation $\frac{x}{5 x} \approx \frac{1}{5}$ on each limit becomes more and more accurate (with $x$ >> $1$, the $1$ becomes insignificant at large $x$).

Since there are five of these limits, you get

$\left(\frac{1}{5}\right) \cdot \left(\frac{1}{5}\right) \cdot \left(\frac{1}{5}\right) \cdot \left(\frac{1}{5}\right) \cdot \left(\frac{1}{5}\right)$

$= \frac{1}{{5}^{5}}$, or $\textcolor{b l u e}{\frac{1}{3125}}$.