Let $f(x)=5x^2-5$ and $\epsilon > 0$ be given. Find a $\delta$ s.t. $0<\delta<1$ so that $|x-1|<\delta$ implies $|f(x) - f(1)| < \epsilon$. For $\epsilon=.2$, find one such $0<\delta<1$.
What is the optimal solution for this problem? Here is my work:
Let $\delta = 1/2$. $|x-1| < 1/2$, so $x+1<7/2$. $|5x^2-5|=5|x+1||x-1|<5(7/2)|x-1|<\epsilon.$ So, $|x-1|<2\epsilon/35$. Finally, $\delta=min\{1/2,2\epsilon/35\}$
When $\epsilon=.2$, $\delta=.0114$.
Here we have $5x^2 - 5 = 0$ when $x = 1$
So for an $\epsilon = 0.2$, we need to solve the inequality $|f(x)-f(1)|<0.2$
$|5x^2 - 5 - 0| < 0.2$
$-0.2 < 5x^2 - 5 < 0.2$
$4.8 < 5x^2 < 5.2$
$.96 < x^2 < 1.04$
$0.9798 < x < 1.0198$
Hence $\delta_1 = .0202$ and $\delta_2 = .0198$
It is the general rule to take the smaller of the two deltas to ensure that both inequalities are met. If $.0202$ were used, then $5(1.0202)^2 - 5 = 0.204 > 0.2$.