# L'Hospital rule, exponental ratio

$$\lim_{x\to ∞} \frac {x^{1000000}} {e^x}$$ could anyone please provide some hits with what result I will end up?

After all applyings of L'Hospital rule, I will get $\frac {n} {e^x}$, where $n$ is large number before I got out of the $x$ powers. So, will it be the limit $0$ then? Since the infinity is nothing I have $\frac {n} {0}.$ Or will it be just the $\infty$?

• "Since the infinity is nothing": what??? – Omnomnomnom Jul 27 '15 at 18:44
• @Omnomnomnom Picky picky picky... heh – David C. Ullrich Jul 27 '15 at 18:45
• @DavidC.Ullrich I honestly don't know what that statement could mean – Omnomnomnom Jul 27 '15 at 18:47
• @Omnomnomnom Sorry. I thought it was sufficiently clear that the statement was nonsense that it would be clear I was just teasing. "what???" is exactly right, even the right number of question marks... – David C. Ullrich Jul 27 '15 at 18:50
• Simply note that $e^x$ grows faster than ANY polynomial function. – Elliot Gorokhovsky Jul 28 '15 at 3:22

The limit will be 0. Another way to see this:

Note that $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots + \frac{x^{1000001}}{1000001!} + \cdots > \frac{x^{1000001}}{1000001!}$$ $$0<\frac{x^{1000000}}{e^x} < \frac{1000001!}{x}$$

$$\lim_{x \to \infty} \frac{1000001!}{x} = 1000001! \lim_{x \to \infty} \frac1x = 0$$

The Squeeze Theorem will give us this result.

Repeated L'Hospital will get you to

$$\lim_{x\to \infty}\frac{10000000!}{e^x}$$

and when $x$ tends to infinity, you'll get a $0$.

• The last step in which you would "differentiate away" the constant is, strictly speaking, an incorrect application of L'Hospital. – Omnomnomnom Jul 27 '15 at 18:51
• That's true @Omnomnomnom, my mistake. – Eemil Wallin Jul 27 '15 at 18:51

A "quicker" approach: note that $$\lim_{x \to \infty} \frac{x^{1000000}}{e^x} = \left(\lim_{x \to \infty}\frac{x}{e^{x/1000000}}\right)^{1000000}$$

• Although, requires you to already know $\lim_{x \to \infty} \frac{x}{e^x} = 0$. – 6005 Jul 27 '15 at 19:47
• @6005 Or you can just use L'Hopital once, instead of 10^6 times in the other approaches. – Teepeemm Jul 27 '15 at 19:58
• @Teepeemm - I've always preferred using it once, rather than 10⁶ times. – Alec Jul 28 '15 at 12:59

The exponential beats any polynomial, so the limit is zero. If you really want to think by L'Hospital, once you differentiate $1000000$ times, the numerator will be a huge fixed number and the denominator will be $e^x$. And $\lim_{x \to +\infty} K/e^x = 0$ for any constant $K$.

If you insist on using L'Hopital, here is how you should think about it.

First remember what L'Hopital says. It says that if you have two functions of $x$, say $f(x)$ and $g(x)$, and you want to know what $f(x)/g(x)$ tends to as $x$ tends to some limit $a$, then if $f(x)$ and $g(x)$ both tend to $0$ or both tend to $\infty$, and if $f'(x)/g'(x)$ tends to a limit as $x$ tends to $a$, then these limits are the same. That is, $$\lim_{x \to a}\frac{f(x)}{g(x)}=\lim_{x \to a}\frac{f'(x)}{g'(x)}$$

provided they both exist. In your case we have $f(x)=x^{1000000}$ and $g(x)=e^{x}$. If we just differentiate once, then the limit of the top and the bottom is still $\infty$, so we can use L'Hopital again. In fact, if you use L'Hopital $1000000$ times, we that $$\lim_{x \to \infty} \frac{x^{1000000}}{e^{x}}=\lim_{x \to \infty}\frac{1000000!}{e^{x}}$$ Where $1000000!$ is the product of all the numbers from $1000000$ to $1$. But this is just a constant divided by $e^{x}$, so when $x$ gets big this large number stays the same, while $e^{x}$ keeps growing. And since $e^{x}$ can get as large as we could ever want, your limit has to be $0$.

This is so easy when you put $e^{x} = t$ and then $t \to \infty$ as $x \to \infty$ so that the expression changes to $f(t) = (\log t)^{a}/t$ where $a = 1000000$. This can be further rewritten as $$f(t) = \left(\frac{\log t}{t^{b}}\right)^{a} = \{g(t)\}^{a}\tag{1}$$ where $b = 1/a = 0.000001$. Now we show that $g(t) \to 0$ as $t \to \infty$ and this will imply that $f(t) \to 0$ as $t \to \infty$.

Let us choose a number $c$ such that $0 < c < b$. Since $t \to \infty$ we can assume $t > 1$ so that $t^{c} > 1$ and hence $$0 < \log t = \log\{(t^{c})^{1/c}\} = \frac{\log t^{c}}{c} \leq \frac{t^{c} - 1}{c} < \frac{t^{c}}{c}\tag{2}$$ where we have used the standard inequality $$\log x \leq x - 1$$ for $x > 1$ with $x = t^{c}$. From $(2)$ and definition of $g(t)$ we get $$0 < g(t) = \frac{\log t}{t^{b}} < \frac{t^{c}}{ct^{b}} = \frac{1}{ct^{b - c}}\tag{3}$$ for $t > 1$. Since $b - c > 0$, it follows that $t^{b - c} \to \infty$ and using Squeeze theorem on equation $(3)$ as $t \to \infty$ we see that $g(t) \to 0$ as $t \to \infty$.