Is there an example of using L'Hospital's Rule on a product where it doesn't work?

I was reading that, when trying to solve something like:

$$\lim_{x\to\infty} f(x)g(x)$$

I can rewrite is as:

$$\lim_{x\to\infty} \frac{f(x)}{\frac{1}{g(x)}}$$

and use L'Hospital's Rule to solve. And, if this doesn't work, I can try using the other function as the denominator:

$$\lim_{x\to\infty} \frac{g(x)}{\frac{1}{f(x)}}$$

So I wondered: are there well-known quotients of functions that don't work in either case and, if so, how do I then solve them?

An example that doesn't submit to this process is:

$$\lim_{x\to\infty} x.x$$

But obviously L'Hospital's Rule would not be necessary in this case.

• Something like $\sqrt{1+x^2}/x$? – David Mitra Aug 22 '12 at 22:47
• Maybe this thread is of interest to you. – t.b. Aug 22 '12 at 22:48
• @AsafKaragila Fixed! :) – Korgan Rivera Aug 22 '12 at 22:56
• @DavidMitra Ha! That example is perfect. – Korgan Rivera Aug 22 '12 at 23:01
• @t.b. Thanks, that's exactly the sort of thing I wanted to see. – Korgan Rivera Aug 22 '12 at 23:04

Functions that have general tendencies to have L'Hopital's not solve anything such as the $e$ function. For example, $$f(x) = \frac{e^{x} - e^{-x}}{e^{-x}+e^{x}}$$ If you were to apply L'Hopital's you would go in a circle of deriving it again and again. Although this problem could be solved with algebra, it still does suit your question.
Usually, $f(x)g(x)$ needs to be converted to a fraction when it is of the indeterminate form: $0\cdot\infty$. Examples:
1) $\lim \limits_{x\to\infty} x\cdot\sin{\frac{1}{x}}=...=1.$
2) $\lim \limits_{x\to\infty} x^2\cdot\sin{\frac{1}{x}}=...=\infty.$
3) $\lim \limits_{x\to\infty} x\cdot\sin{\frac{1}{x^2}}=...=0.$
4) $\lim \limits_{x\to 0} x\cdot\sin{\frac{1}{x}}=...=0.$ (Note: it is not indeterminate, so there is no need to convert to a fraction).