The limit of a rational function

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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...

Imad,

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

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