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

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.)
