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: $\sqrt{-\epsilon+16}<x<\sqrt{\epsilon+16}$
Where do I go from here?
On
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$$|x^2+6-22|<\frac\varepsilon9\times9=\varepsilon.$$
On
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 $x^2\ge0$,
$$\max(0,16-\epsilon)< x^2<16+\epsilon,$$
leading to
$$\sqrt{\max(0,16-\epsilon)}-4< x-4<\sqrt{16+\epsilon}-4.$$
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:
$-1<x-4<1$
Adding $8$ to all sides we obtain $7<x+4<9$, 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:
$|x^2+6-22|<9\dfrac{\epsilon}{9}=\epsilon$