1
$\begingroup$

Prove that if $\lim_{x \to c}g(x)=b$ and $\lim_{x \to b}f(x)=L$ and there exists a sequence $a_n$ converging to the limit $c$ such that $g(a_n)=b$ then prove that $$\lim_{x \to c}f(g(x))$$ does not exists given that $f(b) \neq L$ $$$$If we consider the difference $|f(g(x))-L|$ and if we choose $\epsilon < |L-f(b)|$ then for that $\epsilon$ in every neighbourhood of $c$ we can find $x=a_n$ such that $$|f(g(x))-L|=|f(b)-L|>\epsilon$$ so $$\lim_{x \to c}f(g(x)) \neq L$$ Also if $x$ approaches $c$ by taking any sequence except for $a_n$ then in that case there exists a neighborhood of $c$ such that for $x$ belonging to that neighborhood $g(x) \neq b$ and in that case $$\lim_{x \to c}f(g(x))=L$$ So overall the limit $$\lim_{x \to c}f(g(x))$$ does not exists. $$$$Is My Proof Correct?

$\endgroup$
0
$\begingroup$

Your proof is mostly good, but not correct. But then, neither is the statement you are trying to prove. Here is a counter-example:

  • $c = 0, g(x) = 0$ for all $x$, and $f(x) = \begin{cases}1, &x \ne 0\\0,& x = 0\end{cases}$. And $a_n = \frac 1n$ for all $n$. Then $$\lim_{x\to 0}g(x) = 0\\\lim_{x \to 0} f(x) = 1\\f(0) = 0\\g(a_n) = 0\quad\forall n$$ yet still $\lim_{x \to 0} f(g(x)) = 0$. I.e., it converges.

Where your proof goes wrong is here:

Also if $x$ approaches $c$ by taking any sequence except for $a_n$ then in that case there exists a neighborhood of $c$ such that for $x$ belonging to that neighborhood $g(x) \neq b$ and in that case $$\lim_{x \to c}f(g(x))=L$$

By stating that "any sequence except for $a_n$" would give $L$ as a limit, you assumed that $a_n$ gave every value of $x$ near $c$ with $g(x) = b$, so no other approach to $c$ could have $g = b$. But this was not among the hypotheses, so there is no reason for it to be true.

In order to show that $\lim_{x \to c} f(g(x)) \ne f(b)$, you need at least one means of approaching $c$ where $g \ne b$. However, there is nothing to guarantee this, so the theorem fails.


As an aside about wording, suppose we were also given that $g$ is not constant on any deleted-neighborhood of $c$. Then the theorem would be true, but your wording would still be false:

  • You are still not given that $a_n$ are the only place where $g(x) = b$, so just avoiding $a_n$ is not enough. Instead you could say "for each $n$, let $b_n$ be a point with $|b_n - c| < 1/n$ and $g(b_n) \ne b$."
  • "in that case $\lim_{x \to c}f(g(x))=L$" is wrong. We already know that limit is not $L$. There are no cases about it. What you mean is "in that case $\lim_{n \to \infty} f(g(b_n)) = L$." Respect the symbolism. Never give it your own private meaning, expecting your audience to understand.
$\endgroup$
  • $\begingroup$ But if it is given that $g(x) \neq b$ whenever $x \neq a_n$ then is the proof Correct? $\endgroup$ – user763338 Mar 26 at 3:43
  • $\begingroup$ If that were given, and you fixed the wording, then yes, it would be correct. $\endgroup$ – Paul Sinclair Mar 26 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.