Let $A$ and $x_0$ be real numbers, and let $f(x)$ be a real-valued function defined in a deleted neighborhood of $x_0$. Use the definition of limit to prove that $\lim \limits_{x \to x_0} f(x) = A $ if and only if $\lim \limits_{x \to 0} f(x_0 + x) = A$.

First we prove $\lim \limits_{x \to x_0} f(x)=A \implies \lim \limits_{x \to 0}f(x_0+x)=A$

$\lvert f(x)-A \rvert <\epsilon$, $\forall x$ with $0<\lvert x-x_0 \rvert < \delta$ or $x_0- \delta <x<x_0 + \delta$

$\lvert f(x_0+x)-A \rvert < \epsilon$, $\forall x $ with $0<\lvert x-0 \rvert < \delta$ or $-\delta < x < \delta$

So I want to choose a $\delta$ so that $x_0$ disappears?

I'm having a hard time understanding this process, can somebody give me a walkthrough?


1 Answer 1


Suppose $\displaystyle\lim_{x \to x_0}f(x) = A$. Fix $\epsilon>0$. Then there is a $\delta>0$ such that $|x-x_0|<\delta$ implies $|f(x)-A|<\epsilon$.

Therefore, if $\ |(x+x_0)-x_0|<\delta$, then $|f(x+x_0)-A|<\epsilon$. (I suppose this line is the trick)

That is, if $\ |x|<\delta$, then $|f(x+x_0)-A|<\epsilon$.

This means (by definition) that $\displaystyle\lim_{x \to 0}f(x+x_0) = A$

  • $\begingroup$ For the other way, can I set $x=x-x_0$ and say that if $0<\lvert x-x_0 \rvert < \delta$ then $\lvert f(x) -A \rvert < \epsilon$? $\endgroup$ Commented Nov 8, 2015 at 2:08
  • $\begingroup$ Yes, assume the second limit. If $|x-x_0|<\delta$, then $|f(x_0+(x-x_0))-A|<\epsilon$ $\endgroup$
    – H_R
    Commented Nov 8, 2015 at 3:50

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