Question about limit substitution of non invertible functions

Question

I was working on limit substitutions, they make fair sense and are provable when we are dealing with invertible functions. But I was interested in substituting non-invertible functions like $\sin x$, even though the usefulness of that seems limited.

lets say there exist 2 functions $f(x)$ and $g(t)$ such that

$$t=s(x)$$ and $$f(x) = g(t)$$ Then lets say there exists a limit

$$\lim_{x \to a} t=b$$ and we are trying to describe the approach of the following limit in terms of t. $$\lim_{x \to a} f(x) $$ When trying to express the approach of x to a in terms of t, we need to show that x approaches a only when t approaches b. But since $s(x)$ is non invertible, then obviously $\displaystyle \lim_{t \to b} x$ does not exist. And we cannot express the $x \to a$ in terms of $t$. But since we know $f(x)=g(t)$, intuitively the following should be true because we are not involving $x$, there is no need to extract a value of $x$ from $t$.

$$\lim_{t \to b} g(t) = \lim_{x \to a} f(x) $$

I know this is probably wrong, but can someone give a example where this does not work.

Answer 1

You have $t$ as a function of $x$, and you have $g(t) = g(s(x)) = f(x)$. So $f=g\circ s$.

You have $$\lim_{x\to a}t = \lim_{x\to a}s(x) = b. $$

You are trying to find $$\lim_{x\to a} f(x) = \lim_{x\to a}g(s(x)).$$ You are saying that "intuitively" we should have $$\lim_{x\to a}f(x) = \lim_{t\to b}g(t) = \lim_{s(x)\to b}g(s(x)).$$

Consider $$\begin{align*} s(x) &=\left\{\begin{array}{ll} 1&\text{if }x\neq 0\\ 0&\text{if }x=0. \end{array}\right.\\ g(x) &= \left\{\begin{array}{ll} 3&\text{if }x=0\\ 4&\text{if }x=1\\ 5&\text{otherwise.} \end{array}\right. \end{align*}$$ Then $$g(s(x)) = \left\{\begin{array}{ll} 3 &\text{if }x=0\\ 4 &\text{if }x\neq 0. \end{array}\right.$$ Take $a = 0$ Then $$\lim_{x\to 0}s(x) = 1,\qquad\text{and}\qquad\lim_{x\to 0}g(s(x)) = 4.$$ So $b=1$. However, $$\lim_{t\to 1}g(t)=5.$$

If all the functions are continuous everywhere, however, then things work. If $s$ is continuous everywhere and $g$ is continuous everywhere, then $$\lim_{x\to a}g(s(x)) = g\left(\lim_{x\to b}s(x)\right) = g(b) = \lim_{t\to b}g(t),$$ as desired.