# L’Hopital’s Rule – Day 1

He's pretty swank looking, really.

So, today we started going over L'Hopital's Rule, which is a fiendishly hidden concept that would be pretty useful in Calculus 1, except they don't tell you about it until Calculus 2. Why is it useful? Well, in this case, I DO know why.

L'Hopital's Rule essentially allows you to reduce limits that are in indeterminate form to something more workable--which is saying a lot. But don't trust me--here's an example!

$\large \dpi{150} \lim_{x \to \infty } \frac{5x^3 -2x}{7x^3 + 3}$

Now because I'm so bad at math, I couldn't have told you how to solve that limit a yesterday. Ordinarily, you'd use a direct substitution and, failing that, you could try to factor it. If none of the normal methods work, however, you're sort of dead in the water. L'Hopital's Rule to the rescue!

L'Hopital's Rule basically tells us that if a limit is in indeterminate form (in other words, if direct substitution results in  $\dpi{80} \frac{0}{0}$$\dpi{80} \frac{\infty}{\infty}$ ,  $\dpi{120} 1^{\infty}$$\dpi{120} 0^{0}$,  $\small \dpi{100} \infty ^0$,  $\small \dpi{100} \infty - \infty$,  , or  $\small \dpi{100} 0 * \infty$), you can reduce it by taking the derivative of the top and the derivative of the bottom.

That's it.

So back to the example!

$\large \dpi{150} \lim_{x \to \infty } \frac{5x^3 -2x}{7x^3 + 3}$

Look at that for a second, then try to plug in infinity for x to see if you can do direct substitution. No? Me either, it comes up indeterminate of the form $\dpi{80} \frac{\infty}{\infty}$. No small surprise, as both the numerator AND the denomenator both have cubed powers. So now what? Well, we could try factoring, but if you take a look you can pretty much eyeball that it won't work--that is, if you factor an x out of the top, you're still stuck with a denominator that has multiple terms, so you can't cancel anything. Lame! L'Hopital!

Step 1: Derive the numerator

$\large \inline \dpi{150} \frac{d}{dx} 5x^3 - 2x = 15x^2 - 2$

Step 2: Derive the denominator

$\large \inline \dpi{150} \frac{d}{dx} 7x^3 + 3 = 21x^2$

Now put it all back together:

$\large \dpi{150} \lim_{x\rightarrow \infty} \frac{15x^2 - 2}{21x^2}$

Nice! Now try plugging infinity back in.

Uh oh. Didn't work? DOWN WITH L'HOPITAL'S RULE BURN IT AND HIS STUPID HEAD OF HAI--wait, what? Oh. Well, L'Hopital's Rule can be repeated as many times as necessary, and in fact it IS necessary much of the time. That's less fun, but okay. L'Hopital! Again!

1. Derive the numerator

$\inline \dpi{150} \frac{d}{dx} 15x^2 -2 = 30x$

2. Derive the denominator

$\inline \dpi{150} \frac{d}{dx} 21x^2 = 42x$

Put it all back together:

$\dpi{150} \lim_{x\rightarrow \infty} \frac{30x}{42x}$

Well... when we plug infinity into that, we still get infinity. However, it is clear to see that if we derive both top and bottom again, we will be left with constants! And what does that mean? It means that our limit, after it is simplified, would be a simple

$\dpi{150} \frac{5}{7}$

and you can verify that answer with a graphing calculator, or use the WolframAlpha link in the sidebar to graph the very first iteration of our rational expression. You can see that they are approaching a number. Try graphing the limit to see the exact line that constitutes the limit.

[As a sidenote, I really cannot say enough great things about WolframAlpha. Please go use it.]

Is it really that simple? Yup. Don't get me wrong, when you start throwing in transcendental and trigonometric functions, it gets a lot more complex--and quickly. There are also some exceptions and details I left out, but that is the fundamental explanation of how to execute L'Hopital's Rule on a limit of indeterminate form. I'm not even going to begin with the proof, you can look that up yourself. Duh.

(this is part 1 of a 2 part series--I haven't been to class to learn the second part yet duh)