Let $f$ and $g$ be some functions, assuming all right conditions that allow function composition, I want to prove that $$\lim_{x \to \infty} f(g(x))=f\left(\lim_{x \to \infty}g(x)\right) $$
As long as $\lim_{x \to \infty}g(x)$ exists and it's equal to, let's say, $L$, and $f$ is continuous at $L$. Basically these conditions must be sufficient to guarantee that $\lim_{x \to \infty} f(g(x))$ exists and it's equal to $f(L)$.
So my proof so far goes as follows:
Let $\epsilon>0 $, since $f$ is continuous at $L$, then there exists $\delta$ such that $|f(y)-f(L)|<\epsilon$ provided that $|y-L|<\delta$. Let $y=g(x)$, then provided that $|g(x)-L|<\delta$, it follows that $|f(g(x))-f(L)|<\varepsilon$.
But since $\lim_{x \to \infty}g(x)=L$, for any $\delta>0$ there exists $N\in\Bbb{R}^+$ such that if $x>N$ then $|g(x)-L|<\delta$.
That is, for any $\epsilon>0$, there exists $N\in\Bbb{R}^+$ such that if $x>N$ then it follows that
$$|f(g(x))-f(L)|<\epsilon$$
Which proves that $\lim_{x \to \infty} f(g(x))$ exists and it's equal to $f(L)$. It's very important to me that the proof includes the existence of this limit, since it has great computational value when calculating limits that look a bit "messy".
I'd love some notes on the proof, anything that I might be stating wrong or unclear (writing this kind of proofs is not my forte).
Thanks in advance!
$\endgroup$ 12 Answers
$\begingroup$It is quite well done. Just a few remarks:
- There's no need to write “Let $y=g(x)$”. What comes before together with what comes next is perfectly clear: if $\bigl|g(x)-L\bigr|<\delta$, then $\bigl|f\bigl(g(x)\bigr)-f(L)\bigr|<\epsilon$.
- Why $N\in\mathbb{N}$? The usual definition says that $N\in\mathbb{R}^+$.
Your proof is correct. A short version that does not involve the formal definition of limits would be:
- By the very definition of continuity: $$ \lim_{X\to L} f(X) = f(L) $$ which means is that for any quantity $X \to L$, then $f(X) \to f(L)$.
- By assumption: $$ \lim_{x\to \infty} g(x) = L $$ which means that $ X := g(x) \to L $ as $x \to \infty$
- By substitution you get your result.
