L'Hopital Problem

Pedja

New member
Hint: Use the following identity
$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$

Cubist

Senior Member
I'd apply L'Hôpital to the following...

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

Steven G

Elite Member
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

New member
I'd apply L'Hôpital to the following...

$\lim_{x\to \infty} \frac{\sqrt[3]{\left(1-\frac{4}{x}\right)}-1}{\left(\frac{1}{x}\right)}$
Thanks! How did the (1/x) get in the denominator?

Dr.Peterson

Elite Member
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

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

In small steps...

$\sqrt[3]{\left(x^3-4x^2\right)}-x$$= \sqrt[3]{x^3\left(1-\frac{4}{x}\right)}-x$$= x\left(\sqrt[3]{1-\frac{4}{x}}\right)-x$$= x\left(\sqrt[3]{\left(1-\frac{4}{x}\right)}-1\right)$
$= \frac{x\left(\sqrt[3]{\left(1-\frac{4}{x}\right)}-1\right)}{1}$
divide numerator and denominator by x (which is the same as multiplying both by [imath]\frac{1}{x}[/imath])

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