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 $epsilon = 0.1$. determine $delta$ so that if $|x - a| < delta$, then $|f(x)-l|<epsilon$.

Explanation:

Step1: First find the limit value L

Since $f(x)=x^{2}$ and $x
ightarrow2$, then $L = \lim_{x
ightarrow2}x^{2}=4$.

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

We have $|x^{2}-4|<0.1$. Factor the left - hand side: $|(x - 2)(x + 2)|<0.1$.
When $|x - 2|<\delta$, we assume that $|x-2|<1$ (a common first - step to bound $x + 2$). Then $- 1

Step3: Solve for $\delta$

Since $|(x - 2)(x + 2)|<0.1$ and $|x + 2|<5$ (when $|x - 2|<1$), we have $|x - 2|<\frac{0.1}{|x + 2|}$. To ensure the inequality holds for all $x$ such that $|x - 2|<\delta$, we take the worst - case scenario for the bound of $|x + 2|$. Since $|x + 2|<5$ when $|x - 2|<1$, we solve $|x - 2|<\frac{0.1}{5}=0.02$. Also, we need to consider the condition $|x - 2|<1$. The smaller of the two values of $\delta$ that satisfy the overall condition is $\delta = 0.020$.

Answer:

$0.020$