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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?

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

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  • $\begingroup$ Reference and proof to this? $\endgroup$ – LearningMath Feb 22 '18 at 15:09
  • $\begingroup$ @LearningMath Continuity of $1/x$ isn't enough? $\endgroup$ – egreg Feb 22 '18 at 15:10
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    $\begingroup$ @LearningMath Most calculus texts cover this and some related questions on this site; do a search on reciprocal (rule). You can find a proof of the main limit laws, including this one, on this site. $\endgroup$ – StackTD Feb 22 '18 at 15:12
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    $\begingroup$ +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 $L\neq 0$. $\endgroup$ – MPW Feb 22 '18 at 15:16

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