   SEARCH HOME Math Central Quandaries & Queries  Question from Imad, a student: Dear Sir, I just want to thank you very very much for your help.... I need to find some limits for irrational functions, some of them like this one:               ____ lim       \/ x - 1 - 2 x->5 ---------------- is easy to solve.                  x - 5 But when i found one like this:               3 ____    3 ____ lim         \/ 1 + x - \/ 1 - x x->0 --------------------------------                          x I really try, but i cant so please help me... What you need to recognize here is a difference of cubes

p3 - r3 = (p - r)(p2 - pr + r2)

Thus if p is the cube root of 1 + x and r is the cube root of 1 - x then the numerator of your fraction is the first factor in the difference of cubes expression above. Multiple the numerator and denominator of your fraction by (p2 - pr + r2) to get a numerator of (I am going to use fractional exponents.)

[(1+x)1/3 - (1 - x)1/3][(1 + x)2/3 + (1 + x)(1 - x) + (1 - x)2/3]

= [(1+x)1/3]3 - [(1-x)1/3]3

= (1 + x) - (1 - x)

= 2x

and a denominator of

x [(1 + x)2/3 + (1 + x)(1 - x) + (1 - x)2/3]

Now evaluate the limit of the fraction as x approaches zero.

Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.