Question

visit this desmos link and interact with the graph in order to determine a $delta$ so that if $|x - a| < delta$, then $|f(x)-l|ightarrow2$, and suppose that $varepsilon =.1$. determine $delta$ so that if $|x - a| < delta$, then $|f(x)-l|<varepsilon$.

Explanation:

Step1: Identify the function, limit - point and epsilon

We have $f(x)=x^{2}$, $a = 2$, $L=\lim_{x
ightarrow2}x^{2}=4$ and $\varepsilon=0.1$.

Step2: Set up the inequality $|f(x)-L|\lt\varepsilon$

Substitute $f(x)$ and $L$ into the inequality: $|x^{2}-4|\lt0.1$.
Factor the left - hand side: $|(x - 2)(x + 2)|\lt0.1$.
When $|x - 2|\lt\delta$, we can assume that $x$ is close to 2. Let's first find an upper - bound for $|x + 2|$ when $|x - 2|\lt\delta$. If we assume $\delta\leq1$, then $|x-2|\lt1$ implies $1\lt x\lt3$, and $3\lt x + 2\lt5$. So, $|x + 2|\lt5$ when $\delta\leq1$.

Step3: Rewrite the inequality using the upper - bound

We have $|(x - 2)(x + 2)|\lt|x - 2|\cdot5$ (since $|x + 2|\lt5$ when $\delta\leq1$).
We want $|(x - 2)(x + 2)|\lt0.1$, so $|x - 2|\cdot5\lt0.1$.
Solve for $|x - 2|$: $|x - 2|\lt\frac{0.1}{5}=0.02$.
We also need to check the other part of the double - sided inequality.
We start from $|x^{2}-4|\lt0.1$, which is $- 0.1\lt x^{2}-4\lt0.1$.
$x^{2}-4\lt0.1$ gives $x^{2}\lt4.1$, so $x\lt\sqrt{4.1}\approx2.025$ and $x^{2}-4\gt - 0.1$ gives $x^{2}\gt3.9$, so $x\gt\sqrt{3.9}\approx1.975$.
The distance from $x = 2$ to $x=\sqrt{4.1}$ is $\sqrt{4.1}-2\approx0.025$ and the distance from $x = 2$ to $x=\sqrt{3.9}$ is $2-\sqrt{3.9}\approx0.025$.

Answer:

$0.025$