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Consider $\lim_{x \to 0}(\text{sgn}(x^2))$.

There is an identity for solving limits of composite functions, which states that if $\lim_{x \to a}(f(x)) = b$ and $\lim_{x \to b}(g(x)) = c$, then $\lim_{x \to a}(g(f(x))) = c$.

I want to use this identity to solve $\lim_{x \to 0}(\text{sgn}(x^2))$. And yes, I am aware that there are other ways of solving this limit, but I want to solve it using composite functions. If we set $f(x) = x^2$ and $g(x) = \text{sgn}(x)$, then we have $\lim_{x \to a}(f(x)) = \lim_{x \to 0}(x^2) = 0 = b$. Then $\lim_{x \to b}(g(x)) = \lim_{x \to 0}(\text{sgn}(x))$, which does not exist since sgn is approaching different values as $x$ approaches $0$ from both sides. This means we cannot solve this limit using composite functions so we have to find a different approach.

I am going to make an adjustment to the identity for solving limits of composite functions in order to solve this limit and others like it.

First I am going to take a look at a similar approach using variable substitution, but which actually works. If we let $u = x^2$, then $\lim_{x \to 0}(x^2) = 0$ and $\lim_{u \to 0}(\text{sgn}(u)) = 1$, therefore $\lim_{x \to 0}(\text{sgn}(x^2)) = 1$. The reason it's different this time is because we can take into account the set of possible values of $u$, which is given by the range of $x^2$, which is $[0, \infty)$. There is only one possible direction for $u$ to approach $0$ from, and that is the positive side. So $\lim_{u \to 0}(\text{sgn}(u))$ is essentially the same thing as $\lim_{u \to 0}(\text{sgn}(u) \text{ for } u \ge 0)$, which is essentially the same thing as $\lim_{u \to 0^+}(\text{sgn}(u))$, which is just 1.

Here is how I would fix the identity. If $\lim_{x \to a}(f(x)) = b$ and $\lim_{x \to b}(g\vert_{\text{ran}(f) \; \cap \; \text{dom}(g)}(x)) = c$, then $\lim_{x \to a}(g(f(x))) = c$.

Are there any problems with this identity?

Lauren S
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  • The problem with the identity is that (unless $g$ is continuous at $b$, $\lim_{x \to b} g(x)$ tells you nothing about $g(b)$, So $f(x)$ takes the value $b$ for some $x$'s arbitrarily close (but not equal) to $a$, $\lim_{x \to a} g(f(x))$ might not exist. That is not addressed by your "fix". – Robert Israel May 10 '24 at 18:43
  • The "identity" you quote in the second paragraph is false as stated. Consider $f$ the constant function $0$, and $g(x)$ the function that takes the value $2$ everywhere except at $0$, and $g(0)=3$.Then $\lim_{x\to 1}f(x) =0$, $\lim_{x\to 0}g(x) = 2$, but $\lim_{x\to 1} g(f(x)) = 3$, where your "identity" would say the latter limit "should" be $2$. – Arturo Magidin May 10 '24 at 19:32
  • Whoops I mixed up the operator for intersection and union. I meant to write $g\vert_{\text{ran}(f) ; \cap ; \text{dom}(g)}$ instead of $g\vert_{\text{ran}(f) ; \cup ; \text{dom}(g)}$. I have edited my post. But it looks like the same problem that you've identified is there :/ I will have to think about that. – Lauren S May 10 '24 at 19:33
  • Related: this and this. – Arturo Magidin May 10 '24 at 22:00

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