QuestionFind $\delta$ for $f(x)=x^{\frac{2}{3}}$ as $x \to 1$ with $\varepsilon=0.001$ such that $|f(x)-L|<\varepsilon$ .

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=x^{\frac{}{3}}$ . The limit point is $x \rightarrow1$
3. The given $\varepsilon = .001$
4. We need to find $\delta$ such that if $|x-1|<\delta$ , then $|f(x)-1|<\varepsilon$

STEP 2

First, we need to express the inequality $|f(x)-1|<\varepsilon$ in terms of $x$ and $\delta$ .
$|f(x)-1|<\varepsilon$ $|(x^{\frac{2}{}})-1|<\varepsilon$

STEP 3

Now, we need to express the inequality $|x-1|<\delta$ in terms of $x$ and $\delta$ .
$|x-1|<\delta$

STEP 4

Our goal is to find a relationship between $\delta$ and $\varepsilon$ such that both inequalities hold true. We can start by solving the inequality $|(x^{\frac{2}{3}})-1|<\varepsilon$ for $x$ .

STEP 5

olving the inequality $|(x^{\frac{2}{3}})-1|<\varepsilon$ for $x$ gives us two possible ranges for $x$ .
$- \varepsilon < (x^{\frac{2}{3}})-1 < \varepsilon$ $1 - \varepsilon < x^{\frac{2}{3}} <1 + \varepsilon$

STEP 6

Now, we can take the cube of all parts of the inequality to get rid of the fractional exponent.
$(1 - \varepsilon)^3 < x < (1 + \varepsilon)^3$

STEP 7

Now, we need to find the maximum possible difference between $x$ and $1$ in this range. This will give us our $\delta$ .

STEP 8

The maximum possible difference between $x$ and $1$ in this range is given by the larger of the differences $|1 - (1 - \varepsilon)^3|$ and $|(1 + \varepsilon)^3 -1|$ .

STEP 9

Calculate these differences.
$| - ( - \varepsilon)^3| = | - ( - .001)^3| = | - .999| = .001$ $|( + \varepsilon)^3 -| = |( + .001)^3 -| = |.003 -| = .003$

STEP 10

The larger of these differences is $.003$ , so we set $\delta = .003$ .
Therefore, if $|x-|<.003$ , then $|f(x)-|<.001$ .

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