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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.

Envious Page
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Korgan Rivera
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2 Answers2

1

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.

goldenlinx
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-1

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).

farruhota
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