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question 3
visit this desmos link and interact with the graph in order to determine a $\delta$ so that if $|x - a| < \delta$, then $|f(x) - l| < \varepsilon$ . please make sure your response is accurate to within three decimal places.
let $f(x) = x^{\frac{2}{3}}$, and let $x \to 1$, and suppose that $\varepsilon =.5$.
determine $\delta$ so that if $|x - a| < \delta$ , then $|f(x) - l| < \varepsilon$.
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Explanation:

Step1: Find limit $L$ as $x\to1$

$L = \lim_{x\to1} x^{\frac{2}{3}} = 1^{\frac{2}{3}} = 1$

Step2: Set up the epsilon inequality

We need $|f(x)-L| < \epsilon$, substitute $f(x)=x^{\frac{2}{3}}$, $L=1$, $\epsilon=0.5$:
$|x^{\frac{2}{3}} - 1| < 0.5$
This splits into two inequalities:
$1 - 0.5 < x^{\frac{2}{3}} < 1 + 0.5$
$0.5 < x^{\frac{2}{3}} < 1.5$

Step3: Solve for $x$

Raise all parts to the power of $\frac{3}{2}$:
$0.5^{\frac{3}{2}} < x < 1.5^{\frac{3}{2}}$
Calculate values:
$0.5^{\frac{3}{2}} = \frac{1}{2}\sqrt{\frac{1}{2}} \approx 0.3536$, $1.5^{\frac{3}{2}} = 1.5\sqrt{1.5} \approx 1.8371$

Step4: Find distances to $a=1$

Calculate $|1 - 0.3536| = 0.6464$ and $|1.8371 - 1| = 0.8371$
Choose the smaller distance as $\delta$:
$\delta = \min(0.6464, 0.8371) = 0.646$ (rounded to 3 decimal places)

Answer:

0.646