# If $\lim_{x\to 0}f(x)=L$ then $\lim_{x\to 0}f(cx)=L$ for any nonzero constant $c$.

I was just wondering if this proof is correct.

I'm trying to prove that if $\lim_{x\to 0}f(x)=L$ then $\lim_{x\to 0}f(cx)=L$ for any nonzero constant $c$.

Proof:

If $\lim_{x \to 0}f(cx)=L$ then there exists some $\delta$ such that

$0<|x|<\delta \implies |f(cx)-L|<\epsilon$. We have$|f(cx)-L|=|f(cx)-L+f(x)-f(x)|$. Applying the triangle inequality gives

$|f(cx)-L|=|f(cx)-L+f(x)-f(x)|\leq|f(cx)-f(x)|+|f(x)-L|$. So it suffices to find some $\delta$ such that

$0<|x|<\delta \implies |f(cx)-f(x)|+|f(x)-L|<\epsilon$

Since $\lim_{x \to 0}f(x)=L$ there exists some $\delta_{1}$ such that

$0<|x|<\delta_{1} \implies|f(x)-L|<\epsilon$. Since this must be true for any $\epsilon>0$, it must be true for some $\epsilon>|f(cx)-f(x)|+|f(x)-L|$. So, there exists some $\delta_{1}$ such that

$0<|x|<\delta_{1} \implies |f(x)-L|\leq|f(cx)-f(x)|+|f(x)-L|<\epsilon$.

Letting $\delta=\delta_{1}$ gives the desired

$0<|x|<\delta \implies |f(cx)-f(x)|+|f(x)-L|<\epsilon$.

Please tell me if I did anything invalid. Also I'm new to Calculus so please explain as simply as possible. Thanks

• You go from $|f(x)-L|<\epsilon$ to $\epsilon>|f(cx)-f(x)|+|f(x)-L|$. How do you do this? Jul 31, 2016 at 22:32
• "Since this must be true for any ϵ>0, it must be true for some ϵ>|f(cx)−f(x)|+|f(x)−L|" ... |f(cx) - f(x)| + |f(x)-L| is not a constant and so you can't do this. Your proof fails from here on out. Jul 31, 2016 at 22:34
• the triangle inequality Jul 31, 2016 at 22:35
• You must separately prove that $|f(cx)-f(x)|<\epsilon$. The triangle inequality doesn't allow you to use the same $\epsilon$ in both inequalities. Jul 31, 2016 at 22:37
• And you have to prove it for all epsilon. Not merely some epsilon that is larger than a variable expression based on the function for all values of x. I'm afraid your proof is simply on the wrong track. Jul 31, 2016 at 22:39

This proof is not correct. The key error is when you say

Since this must be true for any $\epsilon>0$, it must be true for some $\epsilon>|f(cx)-f(x)|+|f(x)-L|$.

You're trying to prove that for every $\epsilon>0$, there exists $\delta>0$ such that $0<|x|<\delta \implies |f(cx)-L|<\epsilon$. So you're not allowed to choose whatever $\epsilon$ you want; $\epsilon$ is given to you ahead of time. It is true that for any particular $x$, there exists some $\epsilon$ such that $\epsilon> |f(cx)-f(x)|+|f(x)-L|$. But this isn't any use, since you don't get to choose $\epsilon$. (Moreover, you would need this inequality to hold simultaneously for every $x$ such that $0<|x|<\delta$, and it is not clear how you are getting that.)

• No. As Ross says in his answer, the error is at the very beginning. The proof can’t be patched if it begins by assuming what needs to be proven. The theorem to prove has the form “If this, then that.” There is no valid proof technique for this statement that begins with “If that...” Jul 31, 2016 at 23:05
• @SteveKass: That error is easily fixed, and is more an error of phrasing than of mathematical substance. If you just start the proof with the sentence "So it suffices to prove that..." (and slightly rephrase the part before that to be a justification for this claim), the first part of the proof is fine. Jul 31, 2016 at 23:38

You start with "If $\lim_{x \to 0}f(cx)=L$", but that is what you are trying to prove. You should start with "If $\lim_{x \to 0}f(x)=L$" and prove $\lim_{x \to 0}f(cx)=L.$ I like to think of $\epsilon-\delta$ proofs as challenge and response. If you claim some limit expression, I get to challenge you with an $\epsilon$ and you have to be able to find a $\delta$ that works. Here you are claiming $\lim_{x \to 0}f(cx)=L$, so I get to give you an $\epsilon.$ You are basing your claim on somebody else's claim that $\lim_{x \to 0}f(x)=L$, so you get to give them an $\epsilon'$ and they have to give you a $\delta'$ that supports their claim. The $\epsilon'$ that you choose should be derived from the $\epsilon$ that I gave you. You take their $\delta'$ and derive the $\delta$ that you give me.

• Indeed, that challenge-response model is a great model for all stuff with quantifiers. "For all" means "someone else gets to choose one for you" and "there exists" means "you've got to choose". And of course your goal is to make the statement come true. Aug 1, 2016 at 7:24
• @celtschk: Absolutely. And I always don't understand why it's not taught that way, meaning that a proof corresponds to a winning strategy for the Prover and a disproof corresponds to a winning strategy for the Refuter. This has many ties with deeper aspects of logic that so few students are aware of. Aug 1, 2016 at 10:16
• @ChrisNguyen: See math.stackexchange.com/a/1782071 for an explanation of the challenge-response model that Ross mentioned. Aug 1, 2016 at 10:16