$$\lim_{x\to2}\left(x^2+2x-7\right)\ = 1$$
For every $\epsilon > 0$, there exists a $\delta >0$ such that
$|x-2| < \delta \implies |(x^2+2x-7) - 1| < \epsilon$.
Very often, you solve these problems by looking at what $\epsilon$ needs to do and then working backwards to what $\delta$ needs to do. In this case
$$|(x^2+2x-7) - 1| = |x^2+2x-8| = |(x+4)(x-2)| = |x+4|\,|x-2|$$
So, we need to make $|x+4|\,|x-2| < \epsilon$. We know we are going to make $|x-2| < \delta$, but what do we do with $|x+4|$?
\begin{align}
|x-2| < \delta
&\implies 2-\delta < x < 2 + \delta \\
&\implies 4-\delta < x + 4 < 4 + \delta
\end{align}
The trick is to limit the size of $\delta$. There is no fixed limit that you need to use. Just pick one. I think $10$ is a nice round number so I am going to say, suppose $0 < \delta < 2$. Then
\begin{align}
|x-2| < \delta \; \text{and} \; (0 < \delta < 2)
&\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0<\delta<2) \\
&\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (-2<-\delta<0)\\
&\implies 4 < x + 4 < 8 \\
&\implies |x+4| < 10 \\
&\implies |x+4||x-2| < 10\delta \\
\end{align}
You should see that we now solve $10\delta < \epsilon$ for $\delta$. We get $\delta < \dfrac{\epsilon}{10}$. But wait! We made an assumption that $\delta < 2$. That's very easy to fix. Our final formula is
$\delta = \min\left\{2, \dfrac{\epsilon}{10} \right\}$.
Then we get our proof by adding one more line to the previous argument.
\begin{align}
|x-2| < \delta \; \text{and} \; (0 < \delta < 2)
&\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0 < \delta < 2) \\
&\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (0<\delta<2) \\
&\implies 4 < x + 4 < 8 \\
&\implies |x+4| < 10 \\
&\implies |x+4||x-2| < 10\delta \\
&\implies |(x^2+2x-7) - 1| < \epsilon
\end{align}
Best Answer
First, you can try and prove some important yet simple facts about limits. The following are all equivalent:
$$\eqalign{ & \mathop {\lim }\limits_{x \to a} f\left( x \right) = l \cr & \mathop {\lim }\limits_{h \to 0} f\left( {a + h} \right) = l \cr & \mathop {\lim }\limits_{h \to 0} f\left( {a- h} \right) = l \cr & \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - l} \right] = 0 \cr} $$
You have
$$\mathop {\lim }\limits_{x \to 0} \sqrt {4 - x} + 2 = 4$$
or
$$\mathop {\lim }\limits_{h \to 0} \sqrt {4 - h} + 2 = 4$$
By the above, this is equivalent to
$$\mathop {\lim }\limits_{x \to 4} \sqrt x + 2 = 4$$
So you want to show that for each $\epsilon >0$ there is a $\delta>0$ such that for all $x$, $$0 < \left| {x - 4} \right| < \delta \Rightarrow \left| {\sqrt x - 2} \right| < \varepsilon $$
But multiplying the conjugate gives $$\left| {\sqrt x - 2} \right| = \left| {\frac{{x - 4}}{{\sqrt x + 2}}} \right| < \left| {x - 4} \right|$$
so taking $\delta=\epsilon$ does it.
NOTE You could've also worked with $$\mathop {\lim }\limits_{x \to 0} \sqrt {4 - x} + 2 = 4$$
in fact,
$$\left| {\sqrt {4 - x} + 2 - 4} \right| = \left| {\sqrt {4 - x} - 2} \right| = \left| {\frac{{ - x}}{{\sqrt {4 - x} + 2}}} \right| < \left| x \right|$$
so again, $\delta=\epsilon$, as expected.