For example in this problem: How do I solve delta epsilon proofs for quadratic equations? , one chooses $\delta = 1$ at the beginning, and for this problem: proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$, one chooses $\delta = \frac{1}{2}$. What is the reason for assuming that $\delta$ takes on these values, and how can one see that one should define $\delta$ to be either $1$ or $\frac{1}{2}$, or any other value at the beginning?
Thank you.
You look at the point where you're interested in continuity or the limit, and decide on a radius around it small enough for the graph of the function to be "unproblematic".
A singularity (such as $\frac1{x^2-1}$ has at $\pm 1$) is problematic. Thus, when interested in the limit of that function at $x=0$, we must choose a radius small enough to avoid that. $\frac12$ works.
For a quadratic function, the fact that it eventually overtakes any linear function is problematic. Thus we just need a bound. Any bound works, but $1$ is standard.
You can do without these bounds on $\delta$. But using these bounds means that when choosing a final $\delta$, a linear function of $\varepsilon$ is often enough. That's a lot easier. For instance, take $f(x)=x^2$, and say we want to prove it's continuous at $1$ (or equivalently, that the limit at $1$ is $1$). We can choose $\delta=\min(1,\varepsilon/3)$. Or we can choose $\delta=\sqrt{\varepsilon+1}-1$. The former takes a lot less work to find. Especially when the function gets more complicated.
The core of $\varepsilon$-$\delta$ cares about what happens very close to the point of interest, for very small values of $\varepsilon$. But the way it's conventionally phrased, you unfortunately need to care about what happens when $\varepsilon$ is big as well. Limiting $\delta$ is the easiest way to do that.
It doesn't always work, though. Proving that $\sqrt[3]x$ is continuous at $0$ can't be done by choosing $\delta$ as a linear function of $\varepsilon$, regardless of bounding, as we are on the "problematic" point itself. Here a more complicated expression for $\delta$ is required.
You have to solve the inequation
$$|f(x)-L|<\epsilon$$ or
$$L-\epsilon<f(x)<L+\epsilon.$$
If $f$ is monotonic increasing and invertible, this is equivalent to
$$f^{-1}(L-\epsilon)<x<f^{-1}(L+\epsilon)$$ and you will take $$\delta=\min(x_0-f^{-1}(L-\epsilon), f^{-1}(L+\epsilon)-x_0)$$ to ensure that the interval $(x_0-\delta,x_0+\delta)$ is contained in the inverse image of $(L-\epsilon,L+\epsilon)$.
In other cases, you can simplify the problem by finding an auxiliary function $g$, invertible and such that $|g(x)-L|<|f(x)-L|$.