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$$0<x<\delta\implies\bigl\lvert f(x)-L\bigr\rvert<\varepsilon.$$But then$$-\delta<x<0\implies\bigl\lvert f(-x)-L\bigr\rvert<\varepsilon.$$So, I proved that$$\lim_{x\to0^+}f(x)=L\implies\lim_{x\to0^-}f(x)=L$$and, by the same argument,$$\lim_{x\to0^-}f(x)=L\implies\lim_{x\to0^+}f(x)=L.$$

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – aditya bhatt Oct 19 '19 at 7:26
  • 1
    $\begingroup$ What I wrote is what I meant. If you want more details:\begin{align}-\delta<x<0&\implies0<-x<\delta\\&\implies\bigl\lvert f(-x)-L\bigr\rvert<\varepsilon.\end{align} $\endgroup$ – José Carlos Santos Oct 19 '19 at 7:35
  • $\begingroup$ Sir, thank you for giving me this insight for proof. I used this insight in another problem(problem 34) and could prove the required question and I'm sure it will be useful in many other problems. Thanks a lot! $\endgroup$ – aditya bhatt Oct 19 '19 at 14:42
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Oct 19 '19 at 14:43
  • 1
    $\begingroup$ I suggest no other book. Spivak's Calculus is the best one, as far as I am concerned. $\endgroup$ – José Carlos Santos Oct 19 '19 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.