google L hopital's rule and there's your answer. i'm way too lazy to do it myself lol

The answer is, 0. As recommended earlier in the thread, apply L'Hopitals rule, taking the derivative of the numerator and the denominator. Derivative of the numerator = z/sqrt(5+z^2). Derivative of the denominator = (-1+2z)/2sqrt(-z+z^2). So now after applying L'Hopitals, we have a new limit as z->0 of [ z/sqrt(5+z^2)]/[(-1+2z)/2sqrt(-z+z^2)]. This can be equivalently expressed as [ z/sqrt(5+z^2)] x 1/[(-1+2z)/2sqrt(-z+z^2)]. This simplifies to 2zsqrt(-z+z^2)/sqrt(5+z^2)(-1+z). As z goes to 0, this goes to 0/-sqrt(5) = 0.

TLDR; Use L'Hopitals rule, limit is 0.

Edited for typo.

This post was edited by sunmilk on Jul 6 2012 12:49am

PAY THE MAN

I feel a bit turned on

How much FG does sunmilk get?

This post was edited by Atreyu1502 on Jul 6 2012 12:52am

goodluck btw, my friend studied that for awhile and passed it recently

Did OP finish his homework and go beddy bye?

This post was edited by sunmilk on Jul 6 2012 12:55am

i dont recall Gmat math prob has much of calculus. but maybe im wrong, passed it high over 5 yrs ago

i could do it for you but i have no clue where your squareroots begin or end...

the answer is 3.14