The epsilon delta definition of limits says that if the limit as $x\to a$ of $f(x)$ is L, then for any $\delta>0$, there is an $\epsilon>0$ such that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$.
But the problem is that this definition says very generally that for ANY $\delta$, there is SOME $\epsilon$. So what if I always choose $\epsilon=\infty$? Then it is guaranteed that the distance between $f(x)$ and $L$ is less than $\epsilon$, and, as a bonus, $L$ can literally be anything, which means that the limit can be any value you like. Which is obviously absurd. What am I missing here?
Also, most people say that this definition intuitively tells us that $f(x)$ can be as close to $L$ as you like, because if $\delta$ gets smaller and smaller and approaches zero, then epsilon gets smaller and smaller and approaches zero as well. But this can't be right, as $\epsilon$ is not a function of $\delta$ or something, so you can't say that if one approaches 0, then the other will as well.
Edit: I feel like the problem has to do with the fact that usually when people use this definition to solve limit problems, then they obtain some expression for epsilon as a function of delta (like I write about above), and using this expression, you usually find that as delta goes to zero, then epsilon goes to zero as well. If it was assumed in the definition itself that this should ALWAYS be the case, then the definition would make total sense to me, but it doesn't seem like it does to me. If someone could share some thoughts on this, then I would be very happy.