Functions and inequalities

Question

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 $$|x -3 |< \min\{2,a\} \implies |f(x)-10|<\epsilon.$$

Answer 1

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:

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

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

Answer 2

Assume that $x \neq 3$ to begin with:

$|f(x) - 10| = |x^2 - 10 - 10| = |x^2 - 20| < \epsilon \iff 20 - \epsilon < x^2 < 20 + \epsilon$.

Now let $t = x - 3$, then $x = t + 3$. So:

$20 - \epsilon < (t+3)^2 < 20 + \epsilon$. Thus:

$\sqrt{20 - \epsilon} < |t+3| < \sqrt{20 + \epsilon}$. Thus:

$-\sqrt{20 + \epsilon} < t + 3 < \sqrt{20 + \epsilon}$, and:

$t + 3 < -\sqrt{20 - \epsilon}$ or $t + 3 > \sqrt{20 - \epsilon}$. So:

$-3 - \sqrt{20 + \epsilon} < t < -3 + \sqrt{20 + \epsilon}$, and:

$t < -3 - \sqrt{20 - \epsilon}$ or $t > -3 + \sqrt{20 - \epsilon}$. Thus:

$-3 - \sqrt{20 + \epsilon} < t < -3 - \sqrt{20 - \epsilon}$ or $-3 + \sqrt{20 - \epsilon} < t < -3 + \sqrt{20 + \epsilon}$. So to give one such "positive" $a$ ( $a > 0$ ), we can take:

$a = \text{min} \{-3 + \sqrt{20 + \epsilon},3 - \sqrt{20 - \epsilon}\}$.

Note that it is possible for $a$ to be positive if you have a big enough $\epsilon$.