Find $\lim_\limits{x\to -2}{f(x)}$

Question

Let $f:\mathbb{R}\mapsto{\mathbb{R}}$ be an odd function such that: $$\lim_{x\to 2}{(f(x)-3\cdot x+x^2)}=5$$ Find $\lim_\limits{x\to -2}{f(x)}$, if it exists. (so also prove its existence)

Answer 1

$$\lim_{x\to 2}{(f(x)-3\cdot x+x^2)}=5\\ \lim_{x\to 2}{(f(x)-3*2 +4)}=5\\lim_{x\to 2}{f(x)}=5+2=7\\$$ we know $f(-x)=-f(x)$ as it odd function so $$lim_{x\to -2}f(x)=lim_{x\to +2}f(-x)=\\lim_{x\to +2}(-f(x))=-7$$ limit exist because :if we have odd function ,domain is symmetrical interval .so if we have $|x-2|<\delta$ around $x=2$ then we will have $|-x-2|< \delta\\or\\|x+2|<\delta$ around $x=-2$ to complete this prove $$\forall \varepsilon >0 \exists \delta >0 :|x-2| <\delta \Rightarrow |f(x)-7|<\varepsilon $$ so $$ \forall \varepsilon >0 \exists \delta >0 :|x-(-2)| <\delta \Rightarrow |f(-x)-(-7)|<\varepsilon \\\overset{f(-x)=-f(x)}{\rightarrow}\forall \varepsilon >0 \exists \delta >0 :|x+2| <\delta \Rightarrow |-f(x)-7|<\varepsilon \\ \space \\ \forall \varepsilon >0 \exists \delta >0 :|x+2| <\delta \Rightarrow |f(x)+7|<\varepsilon $$

Answer 2

Hints:

$$\lim_{x \rightarrow -2} f(x) =\lim_{x \rightarrow 2} f(-x)=- \lim_{x \rightarrow 2} f(x).$$

$$\lim_{x \rightarrow 2} f(x)=\lim_{x \rightarrow 2} 3x-x^2 +5.$$