Using limits properties when proving using epsilon-delta definition.

Note: This is extra detailed so you can use it as a template for similar problems.

When trying to use the $\epsilon, \delta$ method to show that

$$\lim_{x\to a}f(x)=L$$

Suppose that for some punctuated interval $I=(a-t)\cup(a+t)$ we can find an upper bound $B$ such

$$ \left|\frac{f(x)-L}{x-a}\right|\le B $$

for $x\in I$. Then for $\epsilon >0$ if we let $\delta =\min\left(t,\frac{\epsilon}{B}\right)$. It will follow that if $\left|\frac{f(x)-L}{x-a}\right|\le B$ and $|x-a|<\delta$, then we can multiply the two inequalities and get

$$ |f(x)-L|=\left|\frac{f(x)-L}{x-a}\right|\cdot|x-a|\le B\cdot\frac{\epsilon}{B}=\epsilon$$

To apply this to $\lim\limits_{x \to 5} \sqrt{2x+6} = 4$ we pick a punctured $t$ interval such as $t=1$ (this value of $t$ is frequently sufficient, but you can pick a smaller $t$ if necessary). So we will try to find an upper bound $B$ for $\left|\dfrac{\sqrt{2x+6}-4}{x-5}\right|$ for $x\in(4,5)\cup(5,6)$.

We can deal with the square root by rationalizing the numerator.

For $x\ne5$

$$ \left|\dfrac{\sqrt{2x+6}-4}{x-5}\right|=\left|\dfrac{\sqrt{2x+6}-4}{x-5}\cdot\frac{\sqrt{2x+6}+4}{\sqrt{2x+6}+4}\right|=\left|\frac{2}{\sqrt{2x+6}+4}\right|$$

Now we want to find an upper bound $B$ on the value of this expression on the interval $(4,6)$ excluding $x=5$.

We will work our way up to the correct inequality in stages.

\begin{eqnarray} 4&<x&<6\\ 8&<2x&<12\\ 14&<2x+6&<18\\ \sqrt{14}&<\sqrt{2x+6}&<\sqrt{18}\\ \sqrt{14}+4&<\sqrt{2x+6}+4&<\sqrt{18}+4 \end{eqnarray} The reciprocals inequalities will have the opposite sense, so we have for $x\in(4,6)$, and $x\ne5$ it is the case that

$$ \left|\frac{2}{\sqrt{2x+6}+4}\right|<\frac{2}{\sqrt{18}+4} $$

So this will do for an upper bound. We can choose a larger upper bound if it makes the problem simpler. For example, it will be a larger upper bound if we use $\sqrt{16}$ instead of $\sqrt{18}$, giving $B=\frac{1}{4}$, then $\frac{\epsilon}{B}=4\epsilon$

Now we are ready to make our $\epsilon, \delta$ proof. Everything up to now is preparation.

Let $\epsilon>0$. Then if $\delta =\min\left(1,4\epsilon\right)$ it follows that if $|x-5|<4\epsilon$

$$ |\sqrt{2x+6}-4|=\left|\frac{\sqrt{2x+6}-4}{x-5}\right|\cdot|x-5|<\frac{1}{4}\cdot4\epsilon=\epsilon $$

There are many ways. For example when $x>0$ we have $$|\sqrt {2x+6}\,-4|=\left|\frac {(\sqrt {2x+6}\,-4)(\sqrt {2x+6}\,+4)}{\sqrt {2x+6}\,+4}\right|=$$ $$=\left|\frac {2x-10}{\sqrt {2x+6}\,+4}\right|\le$$ $$\le \left|\frac {2x-10}{4}\right|=\frac {1}{2}|x-5|.$$ So for $e>0,$ let $d=\min (1, e).$ Then we have $d>0$ and $$|x-5|<d\implies (x>0\land |x-5|<d)\implies$$ $$\implies |\sqrt {2x+6}\,-4|\le \frac {1}{2}|x-5|< \frac {1}{2}d\le \frac {1}{2}e<e.$$