I don't understand the implication statement in a quadratic epsilon delta proof.

Question

If I have $\lim\limits_{x\to2} x^2=4$ and then I use the $\epsilon-\delta$ proof, why is $|x+2| \leq 5$ , even though that is not possible because that would imply that $-5 \leq x+2 \leq 5$ but it is not that, it is $3 \leq x+ 2 \leq 5$? I meant to write less than not less than or equal to

Answer 1

In general, if you make some statement $P,$ and then you say "therefore $Q,$" you are not saying that $P$ and $Q$ are completely equivalent. There might be plenty of ways to satisfy $Q$ while $P$ is still false. What matters is that there is no way to make $P$ true and $Q$ false at the same time. In the examples you've given, we start with $P,$ a statement that is true for $x$ within some interval, and then make a statement $Q$ that is true for $x$ in a larger interval. No problem as long as the smaller interval fits completely inside the larger one.

Now, if someone were to say $3 < x+2 < 5,$ and then say that therefore $\lvert x+2\rvert < 4,$ then you would have a faulty "proof," because part of the interval that makes $3 < x+2 < 5$ also makes $\lvert x+2\rvert < 4$ false, for example if $x=2.$

Answer 2

You claim that $\lim_{x\to 2}x^2=4$ with a $\epsilon-\delta$ argument. Fix $\varepsilon \in (0,1)$. Then it has to exist $\delta=\delta(\varepsilon)>0$ such that $$ |x-2|<\delta \implies |x^2-4|<\varepsilon, $$ i.e., $$ 2-\delta<x<2+\delta \implies 4-\varepsilon<x^2<4+\varepsilon. $$ To this aim, fix $\delta=\varepsilon/5$ (which is $<1$). Then $$ x^2<(2+\delta)^2=4+\delta^2+4\delta \le 4+5\delta = 4+\varepsilon $$ and $$ x^2>(2-\delta)^2=4+\delta^2-4\delta>4-4\delta>4-5\delta=4-\varepsilon. $$

Answer 3

For a given $\epsilon>0$, choose $\delta = \min\{1, ?\}$, where $?$ is to be determined and will depend on $\epsilon$. Then $\mid x-2 \mid < \delta$ implies

$\mid x+2 \mid = \mid x-2 + 4\mid \le \mid x-2 \mid + 4 < 1 + 4 = 5$