#### gabriel.colon32

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- Thread starter gabriel.colon32
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Just plug in a very large number for x, like 100,000, and evaluate the expression. What you get will be very very close to the exact answer. If you get .50004000567, then the answer will (probably) be 0.5

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Thanks! How did the (1/x) get in the denominator?I'd apply L'Hôpital to the following...

[math]\lim_{x\to \infty} \frac{\sqrt[3]{\left(1-\frac{4}{x}\right)}-1}{\left(\frac{1}{x}\right)} [/math]

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Division by 1/x is equivalent to multiplication by ... what?Thanks! How did the (1/x) get in the denominator?

Do you see that Cubist first factored x out of the expression, and then changed that multiplication to a division?

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Thanks! How did the (1/x) get in the denominator?

In small steps...

[math] \sqrt[3]{\left(x^3-4x^2\right)}-x[/math][math] = \sqrt[3]{x^3\left(1-\frac{4}{x}\right)}-x[/math][math] = x\left(\sqrt[3]{1-\frac{4}{x}}\right)-x[/math][math] = x\left(\sqrt[3]{\left(1-\frac{4}{x}\right)}-1\right)[/math]

[math] = \frac{x\left(\sqrt[3]{\left(1-\frac{4}{x}\right)}-1\right)}{1}[/math]

divide numerator and denominator by x (which is the same as multiplying both by [imath]\frac{1}{x}[/imath])

[math]=\frac{\sqrt[3]{\left(1-\frac{4}{x}\right)}-1}{\left(\frac{1}{x}\right)} [/math]