How to prove this limit composition theorem?

Question

If $$\displaystyle \lim_{x \rightarrow c}f(x)=l$$ and $$\displaystyle \lim_{x \rightarrow l}g(x)=L$$ and, in every punctured neighbourhood of $c,\;f(x) \neq l,$

then $\displaystyle \lim_{x \rightarrow c}g(f(x))=L$.

Note that the first two conditions are insufficient for the conclusion to hold. (For a clear example, see the first comment in Is there a better counter-example? (problem involving limit of composition of functions)) But if all three conditions are satisfied, the conclusion may be drawn.

This is because the 3rd condition ensures that near $c$, $gf$ won't be forced to take the value of $g$ at $l$. I.e. $\displaystyle \lim_{x \rightarrow c}gf(x)$ will be equal to $\displaystyle \lim_{x \rightarrow l}g(x)$ instead of necessarily being equal to $g(l)$. (Note that $\displaystyle \lim_{x \rightarrow l}g(x)=g(l)$ iff. $g$ is continuous at $l$.)

So that's my informal explanation for the theorem. But I'm having difficulty writing down a proper proof. Please use the sequential definition of limits (instead of epsilon-delta), if required.

Answer 1

Remember that $\lim_{x\to c}f(x)=L$ iff for every sequence $\{x_n\}_{n\in\omega}$ such that $\lim_{n\to \infty}x_n =c$ and $x_n\neq c$ for all $n$ sufficiently large (i.e. there is some $N\in \omega$ such that for all $n\geq N$, $x_n\neq c$) we have $\lim_{n\to \infty}f(x_n)=L$. That is the sequential definition of limit, note how the sequence has to be different from the point it is approximating for large values of $n$, i.e. it has to approximate it without reaching the point -which is the natural idea of a sequence approximating a point-.

Therefore, we have to show that given any sequence $\{x_n\}_{n\in\omega}$ such that $\lim_{n\to \infty}x_n =c$ and there is some $N\in \omega$, then $$\lim_{n\to\infty}g(f(x_n))=L$$.

So, let $\{x_n\}_{n\in\omega}$ be any sequence approximating $c$ in the above manner. We have that $\{f(x_n)\}_{n\in\omega}$ is a sequence approximating $l$, i.e. with $$\lim_{n\to\infty}f(x_n)=l$$ by the definition of limit. Here is where the third hypothesis is essential. Let $U$ be a neighborhood of $c$ such that when punctured $f$ does not take the value $l$, i.e. for all $x\in U-\{c\}$, $f(x)\neq l$. Since $\{x_n\}$ converges $c$, there is some $M\in \omega$ such that for all $n\geq M$, $x_n\in U$.

Let $A=\max\{N,M\}$, then for all $n\geq A$, $x_n\in U$ because $n\geq M$ and $x_n\neq c$ because $n\geq N$, i.e. $x_n\in U-\{c\}$. Thus, for all $n\geq A$, $f(x_n)\neq l$, and therefore by the definition of limit, $$\lim_{n\to \infty}g(f(x_n))=L$$ since $\{f(x_n)\}_{n\in \omega}$ verifies that $\lim_{n\to\infty}f(x_n)=l$ and that for all $n\geq A$, $f(x_n)\neq l$. This completes the proof.

((By petition.) In order to make it clear, since we know $$\lim_{x\to l}g(x)=L\text{,}$$ we can guarantee (thanks to the sequential definition of limit) that for every sequence $\{y_n\}_{n\in\omega}$ such that $\lim_{n\to\infty}y_n=l$ and such that there is some $C\in\omega$ with the property that for all $n\geq C$, $y_n\neq l$, we have that $$\lim_{n\to \infty}g(y_n)=L$$ Now, let $\{y_n\}_{n\in\omega}$ be given by $y_n=f(x_n)$, then we have shown it has the desired properties and thus $\lim_{n\to \infty}g(y_n)=L$, i.e. $$\lim_{n\to \infty}g(f(x_n))=L$$ when substituting $y_n$ by its value $f(x_n)$.) See how the essential thing was that since $f$ takes not the value $l$ in a punctured neighborhood of $c$ we can guarantee that the sequence $\{f(x_n)\}$ has the desired property, i.e. that is different from $l$ for sufficiently large $n$.