Functions and inequalities I have no idea about this question. Please give a hand whoever can. I am a beginner.
Let $f(x)= x^2 - 10$.
Given $\epsilon > 0$, find the value "$a$"  so that $$|x -3 |< \min\{2,a\} \implies |f(x)-10|<\epsilon.$$

If you're looking for a positive $a$, you cannot always find such values - for e.g. if $x=3$ (which always satisfies $|x-3|< \min\{2, a\}$ ), you have $|f(x)-10| = 11$, so for any $\epsilon < 11$ you have a counterexample irrespective of what $a$ is chosen.
Note: 


*

*Given any $\epsilon$, you can find such an $a$ if $\displaystyle \lim_{x \to 3} f(x) = 10$, which is not the case here.

*If $a$ can be negative, then any such $a$ "works" by a trivial argument. 
