Suppose $$f(x)=\sin(x), \lim_{x→π/2}f(x)=1$$ i.e. $$∀ε>0, ∃δ>0, ~~~\text{s.t. if}~~~|x-π/2|<δ → |\sin(x)-1|<ε$$
Here $L+ε$ is larger than the Max(f), so how do we -graphically or conceptually- find $δ+π/2$?
Also, how do we determine $δ-π/2$, since $L-ε$ has several reflections ($s_{_1}, s_{_2}, s_{_3}$)?
Hoping by understanding the idea above to understand why ε is a function of δ, but NOT the other way.
The aim is to take the given epsilon and find the delta that satisfies the requirements. In this case, we're looking for $\delta$ such that $\frac{\pi}{2} - \delta < x < \frac{\pi}{2} + \delta \implies 1 - \varepsilon < \sin(x) < 1 + \varepsilon$. Or, in other words, you want to find an interval of $x$ values for which $\sin(x)$ sits in that horizontal band you drew.
So graphically, what you can do is:
Draw a rectangle with horizontal sides lying on $y = 1 \pm \varepsilon$ and intersecting the vertical line $x = \frac{\pi}{2}$ such that the graph of $y = \sin (x)$ does not touch either horizontal side.
Find the $x$ coordinates of the vertical sides of the rectangle.
Choose the $x$ value that is closer to $\frac{\pi}{2}$ and use that to determine your value of $\delta$.
I've done so using your graph as a basis - note that I could have extended the rectangle out to meet the specific intersection points you found, which would give the maximum possible $\delta$, but it's not required.
Looking at the graph, we have vertical sides at $x = \frac{\pi}{2} - \delta_1$ and $x = \frac{\pi}{2} + \delta_2$, for some values of $\delta_1$ and $\delta_2$. Then if we let $\delta = \min(\delta_1, \delta_2)$, the interval $x \in (\frac{\pi}{2} - \delta, \frac{\pi}{2} + \delta)$ will definitely sit between the vertical sides. And then for any $x$ on that interval, we know that the graph of $y = \sin(x)$ sits between $y = 1 - \varepsilon$ and $y = 1 + \varepsilon$, which are the horizontal sides of the rectangle, and so we can conclude that $|\sin(x) - 1| < \varepsilon$.