Suppose we have a function $f(x) = 9-x$. Now we know that $\lim\limits_{x\to4} f(x) = 5.$
Using the $\epsilon - \delta$ definition:
$|x-1| < \delta \implies |f(x) - 5| < \epsilon.$
What if we take a wrong limit now, lets say 4 for the given function like $\lim\limits_{x\to4} f(x) = 4$. We know its wrong but the epsilon delta defition still works, and I don't really understand how?
Like so :
$|x-4| < \delta \implies |(9-x-4)| < \epsilon$
$= |5-x| < \epsilon \equiv |x-5| < \epsilon$
Now we know if $\delta$ is $A$ and $\epsilon$ is $B$ then If $x < A \implies x < B$ then $B \leq A$.
Using this:
$|x-4| < \delta$ and hence $|x-4|$ is also less than epsilon. $|x-5| < |x-4|$ is also less than $\epsilon$ for all $x > 0$.
So here we have completed the proof which says to every $|x-x_0| < \delta$ there is a $|f(x) - L| < \epsilon$. Where is my mistake? Because the answer turns out to be wrong.
P.S. Someone please edit this document as I'm on a mobile phone and unable to do it.