# How to solve $\epsilon$−$\delta$ limit with square roots of negative values

I am trying to solve this question: Find an equation for $$\delta$$ for any given $$\epsilon$$ for the limit, $$\lim\limits_{x \to 4} \ x^2 + 6$$

I understand the $$\epsilon$$-$$\delta$$ definition generally but I am hung up on solving this one. As I solve the inequalities I run into an issue where I cannot find a way to solve for $$x$$ and not take the square root of a negative.

Here is where I get to for the epsilon inequality: $$\sqrt{-\epsilon+16}

Where do I go from here?

• I am trying to solve this equation What equation, precisely? Commented Jun 4, 2021 at 20:33
• I think I clarified the question some but I am trying to find an equation for delta. My sticking point is solving | $x^2-16$ | < $\epsilon$ Commented Jun 4, 2021 at 20:39
• $-\epsilon + 16$ is not negative if $\epsilon < 16$. Commented Jun 4, 2021 at 21:33

You want to show the limit is $$22$$. So let $$\epsilon>0$$. We have:

$$|x^2+6-22|=|x^2-16|=|x-4|\cdot |x+4|$$

The term $$|x-4|$$ is bounded by the $$\delta$$ we will choose. So we only need to deal with $$|x+4|$$. Suppose that $$|x-4|<1$$. (this is a fine assumption, because we will require $$\delta$$ to be less than $$1$$ in the end). Then:

$$-1

Adding $$8$$ to all sides we obtain $$7, and so $$|x+4|<9$$. And so in this case:

$$|x^2+6-22|=|x-4|\cdot |x+4|<9|x-4|$$

So now choose $$\delta=\min\{1, \frac{\epsilon}{9}\}$$. If $$|x-4|<\delta$$ then in particular $$|x-4|<1$$ and so the inequality $$|x^2+6-22|<9|x-4|$$ holds. Since $$|x-4|<\frac{\epsilon}{9}$$ we obtain:

$$|x^2+6-22|<9\dfrac{\epsilon}{9}=\epsilon$$

That limit is $$22$$. In order to prove it, using the $$\varepsilon-\delta$$ definition of limit, take $$\varepsilon>0$$. Note that, if $$x\in\Bbb R$$,$$|x^2+6-22|=|x^2-16|=|x-4|\times|x+4|.$$If $$|x-4|<1$$, then $$|x+4|<|x-4|+8<9$$. So, take $$\delta=\min\left\{1,\frac\varepsilon9\right\}$$. Then$$|x-4|<\delta\iff|x-4|<1\text{ and }|x-4|<\frac\varepsilon9.$$So, if $$|x-4|<\delta$$ we have$$|x^2+6-22|<\frac\varepsilon9\times9=\varepsilon.$$

The equation that you want to solve for $$\delta$$ is $$|x-4|<\delta\implies|x^2-16|<\epsilon.$$

The second condition can be rewritten

$$16-\epsilon< x^2<16+\epsilon$$ and as $$x^2\ge0$$,

$$\max(0,16-\epsilon)< x^2<16+\epsilon,$$

$$\sqrt{\max(0,16-\epsilon)}-4< x-4<\sqrt{16+\epsilon}-4.$$