Question :
Given that :$\underset{x\rightarrow0^{+}}{\lim}f(x)=L$ and that $g(x)=f(-x)$
Need to prove that $\underset{x\rightarrow0^{-}}{\lim}g(x)$ exists and define it in values of L
My try to answer : From what is given to me $\forall\varepsilon\exists\delta\quad 0<x-0<\delta\rightarrow|f(x)-L|<\epsilon$ And I understand that I need somehow to prove that $\underset{x\rightarrow0^{-}}{\lim}f(-x)\rightarrow-\delta<x-0<0\rightarrow|f(-x)-\tilde{L}|<\epsilon$
However I am stuck with the f(-x) thing, just do not know how to approach it.
Could use some help :)
You are given the following for some $L \in \Bbb{R}$:
$$(\forall\epsilon > 0)(\exists \delta > 0)(\forall x \in \Bbb{R})(0 < x-0 < \delta \implies \lvert f(x)-L \rvert < \epsilon)$$
You want to prove the following:
$$(\forall\epsilon > 0)(\exists \delta > 0)(\forall x \in \Bbb{R})(-\delta < x-0 < 0 \implies \lvert f(-x)-L \rvert < \epsilon)$$
Now, we want to take the inequality $-\delta < x-0 < 0$ and somehow learn something about $-x$ because $-x$ is what's being passed into $f$ in our conclusion. Let's try taking the negative of the inequality (remember to switch inequality signs):
$$\delta > -x > 0 \rightarrow 0 < -x < \delta$$
Now, we can substitute that last inequality into our conclusion to get a logically equivalent statement:
$$(\forall\epsilon > 0)(\exists \delta > 0)(\forall x \in \Bbb{R})(0 < -x < \delta \implies \lvert f(-x)-L \rvert < \epsilon)$$
If you compare this with the conclusion, you'll see that the conclusion is just the given with $-x$ substituted for $x$. Therefore, since the possible set of values for $x$ (which is $\Bbb{R}$) is closed under negation, this statement obviously follows from the given.