# Limit proof for function that have been manipulated.

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 :)

• Notice that the possible values of $x$ in the second limit are the negative of the possible values of $x$ in the first limit. – Noble Mushtak Mar 30 '16 at 19:26

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.

• Thanks for your answer... !! Tried to rely on what was told by you, will this proof work ? Set $-x=z$ $$\begin{array}{cc} \\ \underset{x\rightarrow0^{+}}{\lim}f(z)=L & \iff\forall\varepsilon>0\exists\delta>0\forall y\in\mathbb{R}\quad0<z<\delta\rightarrow|f(z)-L|<\varepsilon\\ \iff & 0<-x<\delta\rightarrow|f(-x)-L|<\varepsilon\\ \iff & -\delta<x<0\rightarrow|f(-x)-L|<\varepsilon\\ \iff & \underset{x\rightarrow0^{-}}{\lim}f(-x)=L=\underset{x\rightarrow0^{-}}{\lim}g(x) \end{array}$$ – Pavel Penshin Mar 30 '16 at 20:53
• Almost. You have to say that $\underset{z \rightarrow 0^+} \lim f(z)=L$ because the variables under the limit and inside the expression should match. Also, $\forall y \in \Bbb{R}$ should actually be $\forall z \in \Bbb{R}$ because you're using $z$, not $y$. Other than those variable changes, you should be good! – Noble Mushtak Mar 30 '16 at 21:00