# 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?

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.
• But if it is given that $g(x) \neq b$ whenever $x \neq a_n$ then is the proof Correct? – user763338 Mar 26 at 3:43
• If that were given, and you fixed the wording, then yes, it would be correct. – Paul Sinclair Mar 26 at 3:45