Answer
Here, to calculate the limit $\lim\limits_{x \to 4} (x^{2} + 6)$ we can see the function $x^{2} + 6$ is a polynomial and, hence, continuous so the limit can be obtained by simply by replacing $x$ with 4.
So, $\lim\limits_{x \to 4} (x^{2} + 6) = 4^{2} + 6 = 16+ 6 = 22$.
Now, let's find $\delta$ to satisfy $|f(x)- 22|< 0.01$ whenever $|x-4|< \delta$.
Observe $|f(x)- 22|$
$ = |x^{2} + 6 - 22| = |x^{2}-16| = |x^{2}-4^{2}| = |(x-4)(x+4)| = |x-4||x+4|$.
Now, since $|f(x)- 22|< 0.01$ , so,
$|x-4||x+4|< 0.01$
$ \Rightarrow |x-4||x-4|< 0.01$
, or, $|x-4|^{2}< 0.01 \Rightarrow |x-4| < \sqrt{0.01}$.
Thus, we get $\delta = \sqrt{0.01} = 0.1$.
Work Step by Step
To find the limit $\lim\limits_{x \to a} f(x)$ if the function $f(x)$ is continuous such as polynomial, sin function , cos function etc. the limit is obtained by replacement of $x$ with the value at which limit is to be calculated, i.e., $a$. If limit turns out to be $L$.
To determine the value of $\delta$ such that $|f(x)- L|< \epsilon$ whenever $|x-a|< \delta$ we have the following steps:
$\textbf{Step I}$ Find the relation between $|f(x)- L|$ and $|x-a|$ using the property of modulus.
$\textbf{Step II}$ Use $|f(x)- L|< \epsilon$ and replace $|f(x)- L|$ using the value in terms of $|x-a|$ as obtained in step I.
$\textbf{Step III}$ Use the properties of inequality in the inequality obtained from step II to get to the bound for $|x-a|$.
The bound obtained in step III is the required value of $\delta$.
