$\varepsilon$–$\delta$ definition of a limit:
For every $\varepsilon$ $>$ $0$, there exists $\delta$ $>$ $0$ such that whenever $0$ $<$ $|x-c|$ $<$ $\delta$, we have $|f(x)-L|$ $<$ $\varepsilon$.
My doubt arised from the following source:
Prove, using delta and epsilon, that $\lim_{x \to 5} ( 3x^{2} – 1 ) = 74$
Feel free to get the bigger picture: refer to "Examples using a Non-Linear Function", highlighted in blue. Note that the picture above it's only part of the proof, which is defining $\varepsilon$
How To Construct a Delta-Epsilon Proof. (n.d.). Milefoot. Retrieved April 27, 2021, from http://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm
According to what I'ved learned, we must prove that $0 < |x – c| < \delta$ $\implies$ $0 \leq |f(x) – L| < \varepsilon $, not $|x – c| < \delta$ $\implies$ $0 \leq|f(x) – L| < \varepsilon $
Best Answer
Consider this function: $$f(x) = \begin{cases} x+2, & x \ne 1 \\ 0, & x = 1, \end{cases}$$ and suppose we are interested in $$L = \lim_{x \to 1} f(x).$$ It is easy to see that we should have $L = 3$. The epsilon-delta definition says that for any $\epsilon > 0$, there exists $\delta > 0$ such that whenever $0 < |x - 1| < \delta$, then $|f(x) - L| < \epsilon$. And we can see in this case that we do in fact require $0 < |x-1|$, since if not, we could choose $\epsilon = 10^{-100}$, and $x = 1$, but $|f(x) - L| = |0 - 1| = 1 > \epsilon$.
The purpose of $\delta$ is to establish a neighborhood around the limit point $c$ such that whenever $x$ is within a ball of radius $d$ but is not equal to $c$, then the value of the function for all points inside that ball will not deviate from $L$ by more than some chosen $\epsilon$. So in order to exclude the limit point itself, we can require $0 < |x - c| < \delta$, or we can write something to the effect of "for any $\epsilon > 0$, there exists $\delta > 0$ such that whenever $x \ne c$ and $|x - c| < \delta$, then..." But the effect is the same.