finding a delta for a limit

Question

why didn't we but $\pm$ sign when we put the square root ?

And why is the absolute sign around "X" ?

Answer 1

Arthur's answer explains why the inequalities are true. The point is that you are not given a task here such as solving for $x$ in $x^2 = 4$ where $x$ can be any real number. The answer to that problem is indeed $x = \pm\sqrt4,$ because "any real number" includes $x = -2$ and $x = 2,$ both of which satisfy $x^2 = 4.$ But in this exercise you already know $x$ has to be close to $2$ and cannot be anywhere near $-2.$

Note that if we were trying to prove $\lim_{x\to-2} x^2 = 4$ instead of $\lim_{x\to2} x^2 = 4,$ the steps up through $$ \sqrt{4 - \epsilon} < \lvert x\rvert < \sqrt{4 + \epsilon} $$ would be correct as shown in the book, but the next step (replacing $\lvert x\rvert$ with $x$) would not be; we would be looking for values of $x$ near $-2,$ so the correct substitution would be $\lvert x\rvert = -x.$

It all seems more complicated than it needs to be.

I blame the textbook's authors for taking this approach in the first place. They are making it look like part of the work you must do in order to prove a limit is to find the intersection points where the function $f(x)$ enters and exits the horizontal band shown in the figure between $y = 4 - \epsilon$ and $y = 4 + \epsilon,$ when in fact that step is completely unnecessary.

The straightforward approach, I think, is to suppose we have chosen a good value of $\delta$ and see what we can say about the range of values $f(x) = x^2$ can take when $2 - \delta < x < 2 + \delta.$ Just to keep these numbers all positive, let's first assume $\delta < 1.$ (You can put as many restrictions as you like on the value of $\delta$ as long as they do not prevent it from being a positive number; remember that the definition of limit just says "there exists $\delta > 0 \ldots .$") So with $\delta < 1,$ we can square all three parts of the inequality: $$ (2 - \delta)^2 < x^2 < (2 + \delta)^2, $$ and expanding this and regrouping it a bit we have $$ 4 - (4\delta - \delta^2) < x^2 < 4 + (4\delta + \delta^2).$$

If we can manage to get $4\delta + \delta^2 < \epsilon$ then we automatically get $4\delta - \delta^2 < \epsilon$ too, so we just need to focus on the right-hand side. We already assumed $\delta < 1,$ so $$ 4\delta + \delta^2 < 4\delta + 1\delta = 5\delta. $$ So we just need to set $\delta$ to $\frac\epsilon5$ or to $1,$ whichever is less, and we have $$ 0 < \lvert x - 2 \rvert < \delta \implies 4 - \epsilon < x^2 < 4 + \epsilon, $$ as desired. Granted, for small values of $\epsilon$ this method will tend to choose values of $\delta$ that are much smaller than the values I think the book was aiming for; but I remind you once again that the requirement in the proof is merely to show that there exists some value of $\delta$ that makes the implication true. You don't score any extra points for finding the largest possible $\delta$ that works.

And the really nice thing about not looking for the boundaries of the blue rectangle in Figure 2.23 is you can still prove limits even when you have no idea how to find those boundaries exactly (such as when $f(x)$ is a fifth-degree polynomial for which the equation of $x$ on those two boundaries cannot be solved by conventional methods). You just need to make sure you are inside the boundaries.

Answer 2

We begin with he inequality $$ 4-\epsilon<x^2<4+\epsilon $$ We see that all terms are positive (as long as $\epsilon$ is small enough), so square roots of the three sides are well-defined.

The square roots of the left and right sides are written out just as is, while $\sqrt{x^2}$ is the same as $|x|$, so that's what they write as the middle side.