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- 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.
To find the $\delta$, note that $\sin x > \frac2\pi x, \forall x \in (0, \frac\pi2)$.
It is required that $|\sin x - 1| = 1 - \sin x < \epsilon$. Let $1 - \sin x < 1 - \frac2\pi x = \frac2\pi(\frac\pi2-x)< \epsilon$, then select $\delta$ as $\frac{\pi}{2}\epsilon$.
It is not hard to check that $|1-\sin x| < \epsilon, \forall |x-\frac\pi2 | < \frac{\pi}{2}\epsilon$ by the symmetric property of $\sin x$ at $x = \frac\pi2$