on a limit of a composition of functions If $\lim_{x\rightarrow o^{-}} f(x)=a$, $\lim_{x\rightarrow o^{+}} f(x)=b$, and $f(0)=c$ with $a\neq b, a\neq c, b\neq c$, then, what can we say about $\lim_{x\rightarrow o^{-}} f(x^2-x)$ ? Does this limit necessarily exist ? If so, how can we show that by the $\epsilon-\delta$ argument ? Any answer will be appreciated.

hint
$$x^2-x=x(x-1)$$
So, when $ x\to 0^- $,
$x<0 $ and. $ x-1<0 $, thus
$$x^2-x \to 0^+$$
and
$$f(x^2-x)\to b$$
******** * $\epsilon-\delta \; $ proof *****
Given $ \epsilon>0 $,
$$\lim_{x\to 0^+}f(x)=b\implies$$
$$\exists \alpha>0\;:\; (0<x<\alpha \implies $$
$$|f(x)-b|<\epsilon)$$
and
$$\lim_{x\to 0^-}(x^2-x)=0^+\implies$$
$$\exists \delta>0 \;:\; (-\delta<x<0\implies $$
$$0<x^2-x<\alpha)$$
thus
$$\exists \delta>0 \;:\; (-\delta<x<0\implies$$
$$|f(x^2-x)-b|<\epsilon)$$
Which proves that $\lim_{x\to0^-}f(x^2-x)=b$.
