# Is $\infty \times 0$ just $\frac00$?

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?

• 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) – 5xum Feb 17 '17 at 14:33
• – Nosrati Feb 17 '17 at 14:35
• @5xum "which is not true" except when it is – JAB Feb 17 '17 at 17:03
• @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. – TheDayBeforeDawn Feb 17 '17 at 19:16
• @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. – Cameron Williams Feb 17 '17 at 19:17

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: $$\lim_{x\to+\infty} xe^{-x} = \lim_{x\to+\infty} \frac x{e^x} = \lim_{x\to+\infty} \frac1{e^x} = 0$$
• 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. – Euro Micelli Feb 17 '17 at 16:45