L'Hopital Problem

[gabriel.colon32]

 

[Pedja]

Hint: Use the following identity
[math]a-b=\frac{a^3-b^3}{a^2+ab+b^2}[/math]

 

[Cubist]

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]

 

[Steven G]

It seems that you do not have to know how to do the problem (very unfortunate) but just need the answer.
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

 

[gabriel.colon32]

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]

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

 

[Dr.Peterson]

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

Division by 1/x is equivalent to multiplication by ... what?

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

 

[Cubist]

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]