# Problem from Spivak Calculus Chapter 5, Problem 30(1)

To prove
$$\lim_{x\to0^+} f(x)= \lim_{x\to0^-} f(-x)$$

I am trying to make my fundamentals strong in limits which is why I am solving the Spivak's book. Now, this problem is seemingly very simple but I'm stuck in the following way:

Let $$\lim_{x\to0^+} f(x)= L$$. Then according to the definition given in spivak calculus for "Limits from above", given $$\epsilon \gt0$$ there is $$\delta \gt 0$$, such that for all $$x$$ if $$0 \lt x \lt \delta$$ then $$|f(x)-L|\lt\epsilon$$.

Now according to the definition given in spivak calculus for "Limits from below", if, $$\lim_{x\to0^-} f(-x)= L$$, then given $$\epsilon \gt0$$ there is $$\delta' \gt 0$$, such that for all $$x$$ if $$0 \lt -x \lt \delta'$$ then $$|f(-x)-L|\lt\epsilon$$

I tried putting $$y=-x$$ in the definition of "limits from above" but couldn't arrive at the definition for "limits from below".

Any help in solving this problem to clear my concept on limits is appreciated, Thank you.

Suppose that $$\lim_{x\to0^+}f(x)=L$$. Take $$\varepsilon>0$$. Then, as you know, there is a $$\delta>0$$ such that$$0But then$$-\deltaSo, I proved that$$\lim_{x\to0^+}f(x)=L\implies\lim_{x\to0^-}f(x)=L$$and, by the same argument,$$\lim_{x\to0^-}f(x)=L\implies\lim_{x\to0^+}f(x)=L.$$
• Here, at $-\delta<x<0\implies\bigl\lvert f(-x)-L\bigr\rvert<\varepsilon$, do you mean $-\delta<-x<0\implies\bigl\lvert f(-x)-L\bigr\rvert<\varepsilon$? If not, are we allowed to change the independent variable of the function this way? – aditya bhatt Oct 19 '19 at 7:26