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When telling us the algebra of limit for functions we were told that,

Let $f,g :D(\subseteq \mathbb{R}) \to \mathbb{R}$ such that $a$ is a limit point of $D$ and also such that $g(x)\ne 0$ for all $x\in D$ and $\displaystyle\lim_{x\to a}g(x)\ne 0$ then, $$\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{\displaystyle\lim_{x\to a}f(x)}{\displaystyle\lim_{x\to a}g(x)}$$ Assume that the values all limits are finite provided they exist.

Is "for all" (see the bold text) here necessary? For example, if $a\in D$ and $g(a)$ is defined then I don't think that it is necessary that $g(a)\ne 0$. In this case I think that it will be sufficient to say that, $$(g(x)\ne 0\ \forall x\in D\setminus \{a\})\land(\displaystyle\lim_{x\to a}g(x)\ne 0) $$ Am I wrong somewhere?

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It is sufficient that $g(x)$ be nonzero in some open set at $a$. Equivalently in an open interval at $a$. If $g$ is continuous this is equivalent to saying $g(a) \ne 0$. However we must have $\displaystyle\lim _{x \to a} g(x) \ne 0$ or the RHS may not exist. If $g$ is continuous this is equivalent to demanding $g(a) \ne 0$.

Oh, so you say, the equality still holds if the LHS fails to exist as well. But suppose $f(x) = g(x) = x$ for all $x \in D = \mathbb R$ and let $a = 0$. Then $f(x)/g(x) = x/x = 1$ at every point. So the LHS is $\lim _{x \to 0} 1 = 1$ which certainly exists. But the RHS evaluates to $0/0$ which is undefined.

Oh, but you further say, what if we choose to evaluate $0/0$ as $1$ for the purposes of this formula? Suppose now that $f(x) =2x$ and $g(x) = x$ for all $x \in D = \mathbb R$ and let $a = 0$. Then $f(x)/g(x) = 2x/x = 2$ at every point. So the LHS is $\lim _{x \to 0} 2 = 2$. But the RHS again evaluates to $0/0$ which we may evaluate as $1$ as we decided. But then the formula becomes $2=1$ which is not true.

It is possible that if $g$ is not continuous we can have $g(a) = 0$ and the formula still holds. For example let $g(x) = 1$ for all $x \ne 0$ and $g(0)=1$. Again let $f(x) = x$ and $a = 0$. The the function $f(x)/g(x) = x$ when $x \ne 0$ and is undefined at $x=0$. However we can still talk about its limit at $0$, which turns out to be $0$. Moreover $\lim f(x) = 0$ and $\lim g(x) = 1$. So the LHS exaluates to $0/1=0$ which is equal the $RHS$.

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  • $\begingroup$ In the second paragraph, the function $g(x)=x$ doesn't satisfy $\displaystyle\lim_{x\to 0}g(x)=0$ and similar is the case for the third paragraph. Furthermore, when you say "...we must have $g(a)\ne 0$ or the RHS mayn't exist.", do you mean when $g$ is continuous or always? $\endgroup$ – user 170039 Apr 10 '15 at 13:09
  • $\begingroup$ To address the first point: In the second and third paragraphs we do have $\displaystyle \lim_{x \to 0} g(x) = 0$. We defined $g(x) = x$ so by definition $g(x) \to 0$ as $x \to 0$. $\endgroup$ – Daron Apr 10 '15 at 13:52
  • $\begingroup$ Sorry, it was a typo. I meant $\displaystyle\lim_{x\to 0}g(x)\ne 0$. $\endgroup$ – user 170039 Apr 10 '15 at 14:37
  • $\begingroup$ I have chosen $\lim g(x) = 0$ so as to illustrate that $\lim g(x) = 0$ is a necessary assumption to make the formula work. $\endgroup$ – Daron Apr 10 '15 at 15:57

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