# using L'Hopital to compare two exponentials

I have a textbook question that asks to use L'Hopital to show which of the two functions: $e^{0.1 x}$ vs $x^{10}$ is dominant as $x \to \infty$. That is, using: $$\lim_{x \to \infty} \left( \frac {x^ {10} } {e^{0.1 x}} \right) = \lim_{x \to \infty} \left( \frac {10x^ 9 } {0.1e^{0.1 x}} \right) = \lim_{x \to \infty} \left( \frac {90x^ 8 } {0.01e^{0.1 x}} \right) = etc$$

I thought $x^{10}$ was, because the denominator seems to be getting not very large quickly. The textbook disagreed.

Also, by graphing this in a graphic calculator, $x^{10}$ is clearly dominant (becomes nearly vertical very quickly).

However, my maths tutor showed a general case for $x^{n}$ and $e^{k x}$ which clearly proves that $e^{k x}$ is dominant for $n>=1$ and $k>0$. His answer matches the answer in the textbook.

So, the question is: if $e^{0.1 x}$ is more dominant, then it should be able to be shown there is a point of intersection for some large $x$ value ($x >> 1$). ie. $$e^{0.1 x} = x ^ {10}$$

Can anyone help resolve this? (I am first year uni doing calculus)

• Using Maxima, bf_find_root(x^10-exp(x/10),x,1,2); gives 1.0101527198538753273, and more importantly, bf_find_root(x^10-exp(x/10),x,2,1000) gives 647.27751243940046947. – Stop hurting Monica May 2 '14 at 12:38

Basically $e^{ax}=1+ax+(ax)^2/2+...+(ax)^{10}/((10)!)+...=\sum_{n=0}^\infty\frac{(ax)^n}{n!}$

so you will see that eventually $e^{ax}$ is dominant over any polynomial as it has "infinite" degree.

You are correct about the root as well, but the problem is that $e^{0.1x}=x^{10}$ is an implicit equation, i.e. we cannot seperate it out to solve for $x$. So we have to do it numerically, the solutions are here from wolfram alpha: https://www.wolframalpha.com/input/?i=solve+e%5E%280.1x%29%3Dx%5E10

A method to solve this numerically is to write:

$g(x)=e^{0.1x}-x^{10}=0$, so now we are finding the root, and we use newtons method:

Let $x_{n+1}=x_n-\frac{g(x_n)}{g'(x_n)}$, and iteratively we should converge to the root, and that root satisfies $e^{0.1x}-x^{10}=0\Rightarrow e^{0.1x}=x^{10}$.

• Yes, as I said, I have seen the proof...but not the point of intersection where $e^ {kx}$ overtakes $x^ {10}$ – cmroanirgo May 2 '14 at 9:27
• @cmroanirgo yes, I have just put in an edit that addresses that :) – Ellya May 2 '14 at 9:28
• Cool. Thanks for that. Newtons is a good way to solve for it. – cmroanirgo May 2 '14 at 9:31
• no worries, happy to help :) – Ellya May 2 '14 at 9:31

We repeat the l'Hopital rule $10$ times we find $$\lim_{x \to \infty} \frac {x^ {10} } {e^{0.1 x}} =\lim_{x \to \infty} \frac {10! } {(0.1)^{10}e^{0.1 x}}=0$$ so we conclude that $$x^{10}=_\infty o(e^{0.1x})$$

• Yes, as I said, I have seen the proof, and this is exactly the argument taken in the text & by the tutor...but not the point of intersection where $e^ {0.1 x}$ overtakes $x^ {10}$ – cmroanirgo May 2 '14 at 9:28
• If you lokk at the equation $e^{0.1 x} = x ^ {10}$, you can transform it to $x=100 \log(x)$ and the point of intersection you look for is close to $x=647.278$ – Claude Leibovici May 2 '14 at 10:55
• Tenacity: l'Hopital $\times 10$! Nice work! – Namaste May 4 '14 at 11:44

If you do not want to run yourself into the ground with repeated applications of l'Hopitals theorem, invest some transformations before: $$\frac{x^{10}}{e^{0.1\cdot x}}=\left(\frac{x}{e^{0.01\cdot x}}\right)^{10}$$ Comparing $x$ and $e^{0.01⋅x}$ is ten times easier than comparing their tenth power.

Consider the ratios of successive values $\frac{e^{0.1(x+1)}}{e^{0.1x}}$ vs. $\frac{(x+1)^{10}}{x^{10}}$, i.e. $e^{0.1}$ vs. $(1+\frac{1}{x})^{10}$.

In the first case, you multiply each time by a constant. In the second, by a term that goes decreasing. The first function grows faster when $e^{0.1}\gt(1+\frac1x)^{10}$, or $e^{0.01}-1\gt\frac1x$, which occurs at $x\approx100$.