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Suppose that $\lim_{x\to a} f(x) =A$ and $\lim_{x\to a} g(x) =B$ and $B \neq 0$. Then $\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)} = \frac{A}{B}$.

This is really well known and very easy to prove by the regular definition of limits, but I'm dealing with another proof which I think is way more direct! But I'm not sure if a fact at the end of the proof is correct, in fact I don't have a clear idea how to show that fact instead of just saying the conclusion!

So here is my proof:

From the definition, if a function has a limit then that function can be shown as the sum of his limit and an infinitely small function: So $f(x)=A+\alpha(x)$ where $\alpha(x)$ is infinitely small as $x \to a$. On the other hand $g(x)=B+\beta(x)$ where $\beta(x)$ is also an infintely small function as $x \to a$. Now, $\frac{f(x)}{g(x)}=\frac{A+\alpha(x)}{B+\beta(x)}=\frac{A}{B} + \left [\frac{A+\alpha(x)}{B+\beta(x)}-\frac{A}{B} \right]$. So by the same definition used at the beginning we have the desired, where $\left [\frac{A+\alpha(x)}{B+\beta(x)}-\frac{A}{B} \right]$ infinitely small as $x \to a$.

So my question is how to show that the last part is infinitely small(as claimed before)?!

davd
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  • What you are doing is nothing other then the taylor series of order 0, with error terms $\alpha(x),\beta(x)$ (that are not infinitely small, how is that even defined?, but simply 0 at the point a and at most of linear order around that). However to prove such things you already need to know the convergence result you want to prove... – mlk Jun 16 '14 at 18:00
  • An infinitely small function is a function such that $\lim_{x \to a}{\alpha(x)}=0$. @mlk – davd Jun 16 '14 at 18:06

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You have that

$$\displaystyle \frac{A+\alpha(x)}{B+\beta(x)}-\frac{A}{B}=\frac{B\alpha(x)-A\beta(x)}{B^2+B\beta (x)}.$$

Since $\alpha(x),\beta(x)\to 0$ as $x\to a$ we have, by definition of limit:

$\forall \epsilon >0 \exists \delta_1>0 \quad\text{s.t.}\quad 0<|x-a|<\delta_1\implies |\alpha(x)|<\epsilon, $ and $\forall \epsilon >0 \exists \delta_2>0 \quad\text{s.t.}\quad 0<|x-a|<\delta_2\implies |\beta(x)|<\epsilon.$

Taking $\epsilon \le |B|/2$ and $\delta=\min\{\delta_1,\delta_2\}$ we have that

$$\displaystyle \left|\frac{B\alpha(x)-A\beta(x)}{B^2+B\beta (x)}\right|\le \frac{|B||\alpha(x)|+|A||\beta(x)|}{B^2-|B||\beta (x)|}\le \frac{(|A|+|B|)\epsilon}{B^2-\frac{B^2}{2}}=\frac{2(|A|+|B|)}{B^2}\epsilon .$$

I think that now you can conclude.

mfl
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  • Thanks, but why we cannot conclude that $\lim_{x\to a}{\left [ \frac{A+\alpha(x)}{B+\beta(x)}-\frac{A}{B}\right ]}=0$ because $\lim_{x\to a}{\alpha(x)}=0$ and $\lim_{x\to a}{\beta(x)}=0$ ?! @mfl – davd Jun 16 '14 at 18:33
  • Ok this is what you have concluded, so we cannot conclude direct using the rule that is to be proved at the beginning, without showing it with the definition! Ok thanks... – davd Jun 16 '14 at 18:37
  • @mfl Hello , can I ask you something about proving this one? – Maths Survivor Mar 13 '18 at 08:25