Proving limits with epsilon delta While attempting to the prove that a limit is correct, many resources first find delta in terms of epsilon. This step I agree with. But then they say that to complete the proof, one has to "check whether this delta works." But I find this last step redundant. If one first demands that $ |f(x)-L| <\epsilon $ for any $\epsilon >0 $ and then using algebra, manages to find some range centred on the point of interest for any $\epsilon >0 $   , why does one now need to reverse the algebra to complete the proof? Is it kept to formally check whether or not the delta is actually centred around the point of interest ?
 A: It is always a good idea to double check your work. In the case of limits, the challenge is to find a solution of the inequation
$$0<|x-x_0|<\delta\implies |f(x)-L|<\epsilon\tag1$$
for $\delta$. (The solution need not be tight.)
For example, when $f$ is strictly growing (hence invertible)
$$|f(x)-L|<\epsilon\iff f^{-1}(L-\epsilon)<x<f^{-1}(L+\epsilon)$$ and you can take
$$\delta\le\min(x_0-f^{-1}(L-\epsilon),f^{-1}(L+\epsilon)-x_0).$$
But when $f$ is not monotonic or cannot be inverted analytically, the resolution process is more error-prone*. Hence it is good advice to plug the answer back in $(1)$.

As said by @md2perpe, the checking work is enough to make a proof.

*More precisely, you have to find a lower bound and an upper bound of $f$ in $[x_0-\delta,x_0+\delta]$ as a function of $\delta$, and express that the range fits in $[L-\epsilon,L+\epsilon]$.
A: To find $\delta$ is "pre-work". In a proof you don't even need to show how you found $\delta$; you only need to show that it works for the given $\epsilon.$
