Proving Composition of Limits 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?
 A: 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.

