Evaluating limits

**asilvester635** · March 6th 2013, 09:10 AM

Did I do it right? And will i need to find the value of C?

**Plato** · March 6th 2013, 09:45 AM

Originally Posted by asilvester635

Did I do it right? And will i need to find the value of C?

That rationalization is incorrect. It is a cube root not a square root.
You need to use $\left( {\frac{{\sqrt[3]{{{{\left( {1 + cx} \right)}^2}}} + \sqrt[3]{{\left( {1 + cx} \right)}} + 1}}{{\sqrt[3]{{{{\left( {1 + cx} \right)}^2}}} + \sqrt[3]{{\left( {1 + cx} \right)}} + 1}}} \right)$

**asilvester635** · March 6th 2013, 10:05 AM

Do I foil the original numerator with the equation that you gave me?

**Plato** · March 6th 2013, 11:27 AM

Originally Posted by asilvester635

Do I foil the original numerator with the equation that you gave me?

I have no idea what FOIL means. Is it some term from some dumb math-ed class?
Just learn to do basic algebra.

Surely you know that $a^3-b^3=(a-b)(a^2+ab+b^2)~!$

$\left( {\frac{{\sqrt[3]{{1 + cx}} - 1}}{x}} \right)\left( {\frac{{\sqrt[3]{{{{\left( {1 + cx} \right)}^2}}} + \sqrt[3]{{\left( {1 + cx} \right)}} + 1}}{{\sqrt[3]{{{{\left( {1 + cx} \right)}^2}}} + \sqrt[3]{{\left( {1 + cx} \right)}} + 1}}} \right) = \frac{c}{{\sqrt[3]{{{{\left( {1 + cx} \right)}^2}}} + \sqrt[3]{{\left( {1 + cx} \right)}} + 1}}$

**asilvester635** · March 6th 2013, 05:58 PM

Sorry ahh BOSS! Got it from yo mama haha

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