So, for some sequence X, to prove lim(X) = L I have to prove | Xn - L | < ε for all ε > 0 and n > M

So, if Xn is hard to work with, am I really just trying to find a sequence Y such that Yn > Xn, and | Yn - L | < ε ?

My professor was talking about finding a "middle ground" sequence but I tuned him out.

There are different patterns where you can use another sequence Y instead of X.
The best is to post here your exact sequence X, so anybody could give you some tips around it.



Notice this exact phrasing for X to tend towards L :

for all ε > 0, a rank M exists such that | Xn - L | < ε for all n > M

It must be clear that M depends on ε.

This is the best advice.