I have 2 questions when dealing with nonexistent limits:

Q1: If $\lim\limits_{x\to a} f(x)=L\neq 0$ and $\lim\limits_{x\to a}g(x)$ does not exist, then is it correct to conclude that $\lim\limits_{x\to a} f(x)g(x)$ does not exist ? I feel the statement true but fail to prove. I tried using Cauchy test for divergence but it did not work. I appreciate your help.

Q2: Suppose $\lim\limits_{x\to a} f(x)=b$ and $\lim\limits_{y\to b}g(y)$ does not exist. I could find examples in which $\lim\limits_{x\to a} g(f(x))$ exists or does not exist. My question is when does it diverge ? What are conditions needed to imply that $\lim\limits_{x\to a} g(f(x))$ does not exist ?


Q1: Yes, this is valid. To see this, write $h(x)=f(x)g(x)$. If $\lim h(x)$ existed, then since $\lim f(x)$ exists and is not $0$, this would imply $\lim h(x)/f(x)$ exists. But $h(x)/f(x)=g(x)$, and we know $\lim g(x)$ does not exist! Thus $\lim h(x)$ cannot exist.

Q2: You can conclude that $\lim g(f(x))$ does not exist if $f$ is open at $a$. This means that for any open interval $(d,e)$ containing $a$, the image $f((d,e))$ contains an open interval around $b$. Roughly speaking, this says that $f(x)$ takes every possible value near $b$ as $x$ approaches $a$. One more concrete condition that implies $f$ is open at $a$ is that $f$ is continuous on an open interval containing $a$ and has neither a local maximum nor a local minimum at $a$.

  • $\begingroup$ Thank you. The answer for the first question is beautiful, simple and clear. The second answer i need more time thinking about it because you did not give a proof. Anyway, thank you. $\endgroup$ – Bungbu Nov 19 '15 at 4:20
  • $\begingroup$ I can elaborate on the proof if you'd like, but it's a good exercise to think about it yourself if you feel up to it. $\endgroup$ – Eric Wofsey Nov 19 '15 at 4:22
  • $\begingroup$ The assumption $f$ is open at $a$ is needed to ensure that there exists a sequence $\left\{y_n=f(x_n)\right\} \in A \ \left\{b\right\} $ approaching $b$ when $x\to a$. Is that right ? $\endgroup$ – Bungbu Nov 19 '15 at 4:39
  • $\begingroup$ Well, not just that there exists a sequence, but that for any sequence $y_n$ approaching $b$, you can choose a sequence $x_n$ approaching $a$ such that $y_n=f(x_n)$ (at least, for all sufficiently large $n$). $\endgroup$ – Eric Wofsey Nov 19 '15 at 4:40

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