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The question is “Prove that the limit as x approaches zero of f(x)=the limit as x approaches a of f(x-a).

I have taken a few stabs at the problem and was told that the solution was the image I attached.

The given proof largely makes sense to me, only it seems to be a proof that the limit as x approaches a of f(x) equals the limit as y approaches zero of g(y-a). I don’t see how this is a proof of what was asked, and I also don’t see how to proceed in the other direction. I think my issue is that I’m struggling to connect my intuitive/verbal understanding with epsilon-delta limit notation, and changing the variable from x to y also confuses me. Can anyone help make sense of this?

Solution

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$$\lim_{x\to 0}f(x)=L\iff$$ $$ \forall e>0\,\exists d>0\, \forall x\,(0<|x|<d\implies |f(x)-L|<e)\iff$$ $$ \forall e>0\, \exists d>0\,\forall x'\,(0<|x'-a|<d\implies |f((x'-a))-L|<e) \iff$$ $$ \lim_{x'\to a}f(x'-a)=L \iff$$ $$ \lim_{x\to a}f(x-a)=L.$$ Regardless of whether or not $a$ is a $0$ of $f.$

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  • $\begingroup$ Ah, thank you. So my confusion was due to the fact that the proof was more general than I was expecting. Also, I’m new here; how do I post with nice notation like you? $\endgroup$ – Samuel Oct 20 '19 at 1:45
  • $\begingroup$ On the top bar, right, click on Help. Choose the Help Center. Click on "How Do I Format Mathematics Here?" Read it and its linked Quick Tutorial. $\endgroup$ – DanielWainfleet Oct 20 '19 at 7:24
  • $\begingroup$ You can also look at MathJax.Org. This site has MathJax in it. E.g. typing a dollar sign before and after \lim_{x\to 0}f(x)=L\iff gives $\lim_{x\to 0}f(x)=L\iff$. $\endgroup$ – DanielWainfleet Oct 20 '19 at 7:33
  • $\begingroup$ As a general rule of thumb, you should not 'answer' with a question. This would be more suited for a comment. $\endgroup$ – MathsofData Mar 26 at 5:23
  • $\begingroup$ @MathsofData: This was posted as an answer last year, but replaced by an unrelated question by a new user. Unfortunately, two users approved this edit suggestion. $\endgroup$ – Martin R Mar 26 at 7:46

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