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Chapter 4: Problem 36

Evaluate the following limits. $$\lim _{x \rightarrow 2} \frac{\sqrt[3]{3 x+2}-2}{x-2}$$

Short Answer

Expert verified
Answer: The limit of the function is $\frac{3}{4}$.

Step by step solution

01

Identify the conjugate of the numerator

The conjugate of the numerator is given by the expression $$(\sqrt[3]{3x + 2} + 2)$$ We will use this expression to manipulate the function.
02

Multiply the function by the conjugate

To multiply the function by the conjugate, we multiply the numerator and the denominator by the conjugate of the numerator. $$\lim_{x \rightarrow 2}{\frac{\sqrt[3]{3x + 2} - 2}{x - 2}} \times \frac{\sqrt[3]{3x + 2} + 2}{\sqrt[3]{3x + 2} + 2}$$
03

Simplify the expression

Now, simplify the expression by multiplying the terms in the numerator and the denominator. $$\lim_{x \rightarrow 2}{\frac{(\sqrt[3]{3x + 2} - 2)(\sqrt[3]{3x + 2} + 2)}{(x - 2)(\sqrt[3]{3x + 2} + 2)}}$$
04

Expand the numerator and simplify further

In this step, expand the terms in the numerator and simplify the expression using the difference of cubes formula. $$\lim_{x \rightarrow 2}{\frac{(3x + 2) - 8}{(x - 2)(\sqrt[3]{3x + 2} + 2)}}$$ We simplify the numerator: $$\lim_{x \rightarrow 2}{\frac{3x - 6}{(x - 2)(\sqrt[3]{3x + 2} + 2)}}$$
05

Factor out the common term in the numerator

We can factor out the common term, 3, in the numerator. $$\lim_{x \rightarrow 2}{\frac{3(x - 2)}{(x - 2)(\sqrt[3]{3x + 2} + 2)}}$$
06

Cancel the common factors

Cancel the common factors, (x - 2), in the numerator and the denominator. $$\lim_{x \rightarrow 2}{\frac{3}{\sqrt[3]{3x + 2} + 2}}$$
07

Evaluate the limit

Now that we have eliminated the indeterminate form, evaluate the limit by substituting x = 2 into the expression. $$\lim_{x \rightarrow 2}{\frac{3}{\sqrt[3]{3x + 2} + 2}} = \frac{3}{\sqrt[3]{3(2) + 2} + 2} = \frac{3}{2 + 2}$$ The limit is: $$\lim_{x \rightarrow 2}{\frac{\sqrt[3]{3x + 2} - 2}{x - 2}} = \frac{3}{4}$$

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