[Math] Composite functions, continuity, and limits

calculuscontinuitydefinitionfunctionslimits

I am looking at a scenario where we take the limit of a composite/nested function

$$\lim _{x \rightarrow c} f(g(x))$$

However apparently there are two jumps being made here that I don't quite understand:

  1. We can rephrase the problem as solving for $f( \lim _{x \rightarrow c} g(x))$. I am sure it's correct and fine but I see this statement being thrown around everywhere with no proof or intuition. Why are we allowed to do this?

  2. Assume the answer to the limit of $g(x)$ is a constant $d$. Then supposedly $f(d)$ has to be continuous at $x=d$. I think this means $f(d)$ is defined and the two-sided limit $L$ at $x=d$ actually equals $f(d)$ as well. Is this right? Why does it have to be continuous at $d$?
    What about the inner function $g(x)$? Does this mean endpoints of a domain are not continuous (I assume they are not)? What if they are defined? For instance a function with domain $(-4, 4)$ versus one with domain $[-4, 4]$?

Best Answer

It may be the case that $f(x)$ and $g(x)$ are discontinuous functions, but their composition $f\circ g(x)$ is continuous.

i.e. lets pick two of the craziest (and most famous) discontinuous functions.

Let $g(x)$ be Thome function

$g(x) = \begin{cases} \frac {1}{q} & x \text{ is rational with } x = \frac pq \text{ in lowest terms}\\0& x\text{ is irrational}\end{cases}$

let $f(x)$ be the Dirichlet function

$g(x) = \begin{cases} 0 & x \text{ is rational with }\\1& x\text{ is irrational}\end{cases}$

$f\circ g(x)$ is continous. But this is pretty unusual.

However if $g(x)$ is continuous at $c$ (as you say above, left hand limit at $c$ equals the right hand limit at $c$ equals $g(c)$) and $f(x)$ is continuous in the neighborhood of $g(c)$ then $f\circ g(x)$ is continuous at $c$ and $\lim_\limits{x\to c} f\circ g(x)$ exists (both left hand and right hand) and equals $f\circ g(c)$

What matters is continuity around $c$ (for $g$) and around $g(c)$ (for $f$). $f(x)$ and $g(x)$ might do crazy stuff outside of those neighborhoods.

Why are we allowed to do this.

We need to go to the $\epsilon-\delta$ definition of limits.

$g$ is continuous at $c$

$\lim_\limits{x\to c} g(x) = L$

$\forall \epsilon_1>0, \exists \delta_1>0: |x-c|<\delta_1\implies |g(x) - L|<\epsilon_1\tag{1}$

And $f$ is continuous at $L$

$$\forall \epsilon_2>0, \exists \delta_2>0: |x-L|<\delta_2\implies |f(x) - f(L)|<\epsilon_2$$

Setting $\epsilon_1 < \delta_2$ then $|f\circ g(x) - f\circ g(c)|<\epsilon_2$

We can do this because 1) is true for any $\epsilon.$

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