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When I solve some limit I get infinity times zero in the answer, but isn't infinity just $\frac10$ and $\frac10 \times 0 = \frac00$. Can I just use L'Hospital's rule there?

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    Infinity is not a number, and dividing by zero is not an algebraic operation. You need to be **very** careful when saying things like $\infty=\frac10$ (which is not true) – 2017-02-17
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    See: https://en.wikipedia.org/wiki/Indeterminate_form – 2017-02-17
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    @5xum "which is not true" [except when it is](https://en.wikipedia.org/wiki/Riemann_sphere) – 2017-02-17
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    @JAB In the field $\mathbb{R}$, expressions like $\frac{1}{0}$ are not well defined. Period. In addition, infinity is not a (real) number. Discussing the Riemann sphere in this context will simply obscure matters for students who are often already confused. – 2017-02-17
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    @JAB even then, it isn't really correct to call it $\frac{1}{0}$ more so than a symbol which represents dividing *any* nonzero number by zero. It might as well be $\frac{2}{0}$ or $\frac{-1}{0}$. That is to say that it is more like an equivalence class. – 2017-02-17
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    @CameronWilliams Point accepted – 2017-02-17

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Can I just use l'Hospitals rule there?

If direct substitution into a limit gives you $\infty \cdot 0$ then you can use l'Hopital's rule but it's a requirement that you must first modify the expression so that direct substitution gives you $0/0$ or $\infty/\infty$.

For a very simple example, consider $\displaystyle\lim_{x \to +\infty} x e^{-x}$. Direct substitution gives $\infty \cdot 0$. But in order to use l'Hopital's rule you must first rewrite: $$ \lim_{x\to+\infty} xe^{-x} = \lim_{x\to+\infty} \frac x{e^x}$$ Now direct substitution gives you $\infty/\infty$, so now you can use l'Hopital's rule: $$ \lim_{x\to+\infty} xe^{-x} = \lim_{x\to+\infty} \frac x{e^x} = \lim_{x\to+\infty} \frac1{e^x} = 0$$

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    To wit, L'Hôpital's rule is very specific about $(num)'/(den)'$. It does not say $(fac1)'*(fac2)'$. It wouldn't even make sense to apply it directly to a $0*\infty$ Situation. – 2017-02-17