L’Hopital’s rule.

Two photos from a recent trip up north. Major bonus points for knowing which of New England's many trails this was taken on.

L’Hopital’s rule is really how every student of calculus (and I believe Leibniz, though I cannot find a reference) wishes the quotient rule worked.  Specifically, that

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}.

Of course, it can’t be that easy.  We also need that f and g are differentiable in a neighborhood of a, that both function approach 0, or they both approach \infty, or they both approach -\infty as x approaches this point a, and finally that the limit on the right hand side exists (though we all recall that if it does not work the first time, we may continue to apply L’Hopital until the limit does exist, which then justifies using L’Hopital in the first place).

I was thinking of this rule in relation to generating interesting examples of limits.  In particular, if we are in a situation where L’Hopital’s applies, then we can apply the rule in two ways:

\lim_{x\to a}\frac{f'(x)}{g'(x)}=\lim_{x \to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{\left(\frac{1}{g(x)}\right)'}{\left(\frac{1}{f(x)}\right)'}.

Proceeding informally (i.e., I’m not going to keep track of hypotheses), the right hand side of this evaluates to
\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)^2}{g(x)^2}\frac{g'(x)}{f'(x)}.

This is all well and good- the right hand side looks appropriately ugly, but now the trick is picking f and g to get interesting limits.  I have worked out two reasonable examples:

1. Choosing f(x) = sin(x)g(x) = x, we get

\lim_{x \to 0} \frac{\sin{x}}{x^2}\tan{x} = 1.

Also, moderate amounts of bonus points for naming (at least) two universities in the northeast with this mascot.

2. Choosing f(x) = e^x-1 and g(x) = \log{x}, and applying (hopefully correctly!) a number of logarithm rules, we can get

\lim_{x \to 1} \frac{(e^x-1)^2}{\log{(x^{\log{(xe^x)}})}} = 0.

What would be interesting is to find an example where it is difficult/impossible to evaluate without recognizing that it was created using this process.  This second example might fit the “difficult” bill, as I would not want to take the derivative of the denominator directly, but factoring, you might recognize it as xe^x (\log{x})^2, and then be able to reverse engineer this process, somehow.

As usual, just a thought I’ve been playing with.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s