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 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.
