I was checking this link Proving that an additive function $f$ is continuous if it is continuous at a single point and both Jspecter and Alex Becker's solutions seem to rely on the fact that $$\lim_{x\rightarrow c} f(x) = \lim_{x\rightarrow a} f(x-a+c)$$ Could someone please explain to me how is that you can change the scalar that x is approaching and still hold the equality? And how can you tell that $$f(x-a+c)$$ is defined? I'm sorry I didn't ask this question in the original post but I don't have the right to comment right now.

Thank you!


This is just substituting the variables - instead of $x\to c$ take $x=y-a+c$, or $y=x+a-c$. Then $y\to a$, and we get that $$\lim_{x\to c}f\left(x\right)=\lim_{x\to c}f\left(x+a-c-a+c\right)=\lim_{y\to a}f\left(y-a+c\right)$$ But instead of renaming the variable as $y$, they kept denoting it $x$.

  • $\begingroup$ Thanks, from what I understand from your answer is that because we know that $$f(a)$$ is defined, then they're adding and subtracting the same term right? But where does the c come from on the right hand side of $$x=y-a+c$$? I know that x was approaching c originally, but isn't adding that c assume that f(c) is defined, which is precisely what we're trying to prove? $\endgroup$ – user280809 Oct 18 '15 at 2:06
  • $\begingroup$ We are trying to prove that assuming $f\left(c\right)$ exists, then $$\lim_{x\to c}f\left(x\right)=f\left(c\right)$$ If $f\left(c\right)$ is not defined, the question whether the function is continuous at $c$ is meaningless $\endgroup$ – Guy Oct 18 '15 at 10:08

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