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### Limit Exercises Worked

Exercises from public domain textbooks. Solutions where provided do not necessary follow the same lines as the original.

#### X-1. {$$\lim_{x \rightarrow 0} {{(x + 1)^{1 \over 3} - 1} \over x}$$}

Method

The approach is to solve this using the difference of two cubes. The formula for the factors of the difference of two cubes is:

{$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$} and we will proceed with

{\begin{align} a &= (x + 1)^{1 \over 3}, \text{ and} \cr b &= 1 \end{align}}

so we multiply the numerator and denominator by the other factor.

Working

{\begin{align} \lim_{x \rightarrow 0} &{{(x + 1)^{1 \over 3} - 1} \over x} \cr &= \lim_{x \rightarrow 0}{{\left( (x+1)^{1 \over 3} -1 \right)\left( (x+1)^{2 \over 3} + (x+1)^{1 \over 3}(1) + 1^2 \right) } \over {x \left( (x+1)^{2 \over 3} + (x+1)^{1 \over 3}(1) + 1^2 \right) }} \cr &= \lim_{x \rightarrow 0}{{ (x+1) -1 } \over {x \left( (x+1)^{2 \over 3} + (x+1)^{1 \over 3}(1) + 1^2 \right) }} \cr &= \lim_{x \rightarrow 0}{x \over {x \left( (x+1)^{2 \over 3} + (x+1)^{1 \over 3}(1) + 1^2 \right) }} \cr &= \lim_{x \rightarrow 0}{1 \over { (x+1)^{2 \over 3} + (x+1)^{1 \over 3}(1) + 1^2 }} \cr &= {1 \over { (0+1)^{2 \over 3} + (0+1)^{1 \over 3}(1) + 1^2 }} \cr &= {1 \over 3} \end{align}}

{$$\color{red}{ {{(x + 1)^{1 \over 3} - 1} \over x} }$$}

Graphs:

1. Desmos Graphing Calculator

Sources:

Recommended:

Category: Math Precalculus Limits

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### August 05, 2017

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