# Inverting a fraction in a limit

I've seen in a limit problem(limit of a fraction) that the fraction can be inverted and then proceed on the new fraction. Is this ok? Can someone show a proof?

• Could you give an example of what exactly happens? – StackTD Feb 22 '18 at 14:56
• Here in the answer of Marcus Scheuer: math.stackexchange.com/questions/1137451/… – LearningMath Feb 22 '18 at 14:57
• But he inverts the limits he finds as well; that's a consequence of a property (see below). – StackTD Feb 22 '18 at 15:03

If you have: $$\lim_{x \to a} f(x) = L$$ with $L \ne 0$, then you also have: $$\lim_{x \to a} \frac{1}{f(x)} = \frac{1}{L}$$ So yes: you can consider the inverted fraction, but then you get the reciprocal of the (non-zero) limit as well.
• @LearningMath Continuity of $1/x$ isn't enough? – egreg Feb 22 '18 at 15:10
• +1. Note this is a consequence of the following. If $\lim\limits_{x \to a} f(x) = L$ and $g$ is continuous at $L$, then $\lim\limits_{x \to a} g(f(x)) = g(L)$. In your case, you are taking $g(x) = 1/x$ so $g$ is continuous there as long as $L\neq 0$. – MPW Feb 22 '18 at 15:16