A student asks:

I know that $x^3 e^{-x}$ approaches zero as $x$ approaches infinity - I can see it from the graph - but I don't really understand why? Can you help?

Of course I can! However, it's going to take us into the murky depths of analysis, and we'll need to understand the idea of a *limit*.

When you're talking about real numbers (and in most other places - although real numbers are easiest to think about), a limit is something you can get as close as you like to, without necessarily reaching it.

You'll have seen this in C2 geometric series: we talk very loosely about the sum to infinity of things like $1 + \frac{1}{2} + \frac{1}{4} + ...$. With any finite number of terms, you can get close to 2, but you can never quite reach it. In fact, you can get as close as you like: if you say "I need to be within $10^{-6}$ of 2," I can tell you how many terms you need to take to get that close.

I know $S_n = \frac{a(1-r^n)}{1-r}$, and in this case $a=1$ and $r = \frac{1}{2}$. That gives $S_n = \frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}}$, and I want to be within a millionth of 2:

$\left| S_n - 2 \right| < 10^{-6}$, and I need to solve for $n$. $\left|\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}} - 2\right| < 10^{-6}$ Let's tidy up that fraction: $\left| 2 - 2 \left(\frac{1}{2}\right)^n - 2\right| < 10^{-6}$ Fix those ||s! $ 2 \left(\frac{1}{2}\right)^n < 10^{-6}$ $\left(\frac{1}{2}\right)^n < 5 \times 10^{-7}$ Now take logs: $ n \log\left(\frac {1}{2} \right) < \log \left(5 \times 10^{-7}\right) $ Divide both sides by $\log\left( \frac{1}{2} \right)$, remembering that that's negative, so the inequality flips: $ n > \frac{\log \left(5 \times 10^{-7}\right)}{\log\left(\frac {1}{2} \right)}$

We can even sort that out a bit, using $\log_{10}$ for the sake of simplicity:

$ n > \frac{ \log(5) - 7}{ - \log(2)} = \frac{7 - \log(5)}{\log(2)}$

Then we call the Mathematical Ninja1, who says $\log_{10}(2) \simeq 0.3$ and $\log_{10}(5) \simeq 0.7$.

$ n \gtrsim \frac{ 7 - 0.7 }{0.3} = 21$

As long as $n$ is at least 21, our sum is within a millionth of 2. A more general proof would use $\delta$ in place of $10^{-6}$ - we could come up with an explicit $n$, in terms of $\delta$, that guarantees we'll be within $\delta$ after we take $n$ terms.

Phew.

A little aside: there's nothing to say we have to find the *smallest* $n$ that works - we just need to find an $n$ such that, if we go beyond it, we're always within $\delta$ of our target.

Well, we're going to do roughly the same thing again: if you give me a small number, $\delta$, I need to find an $n$ such that, if $x > n$, $\left |x^3 e^{-x} - 0\right| < \delta$.2 Let's see how far we need to go - and we know that $x>0$, so we can get rid of the absolute value signs:

$x^3 e^{-x} < \delta$
Now we've got a problem: we can't solve that explicitly. However, if we can find a function that's always bigger that $x^3 e^{-x}$ but which *also* goes to 0, our function must go to zero as well. So how do we find that function?

How about we rewrite this as: $\frac{x^3}{e^{x}}$. We can express $e^{x}$ as $1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + ...$. In particular, $e^{x} > \frac{1}{4!}x^4$ for any positive $x$.

That means: $\frac{x^3}{e^x} < \frac{x^3}{x^4 / 4!} = \frac{24}{x}$. That's a lot easier to work with! $\frac{24}{x} < \delta$ as long as $x > \frac{24}{\delta}$, or more fully:

If $x > \frac{24}{\delta}$, then $\delta > \frac{24}{x} > x^3 e^{-x} > 0$ - which means we can make $x^3 e^{-x}$ as small as we like with a large enough $x$.

That means we're allowed to write: $x^3 e^{-x} \rightarrow 0$ as $x \rightarrow \infty$.

## StuNattrass

RT @icecolbeveridge: [FCM] Why does $x^3 e^{-x}$ go to zero as $x$ goes to infinity?: http://t.co/zFLo8NLfKt

## tkleinwalsh

RT @icecolbeveridge: [FCM] Why does $x^3 e^{-x}$ go to zero as $x$ goes to infinity?: http://t.co/zFLo8NLfKt