Question

How would you go about solving the following problem? lim x->0 [(x+1)^(1/3)-1]/x

Answer 1

Okay, let's give this a shot.

first, let's look at the numerator:

(x+1)^(1/3) - 1

We are going to look at this as one of the factors of the difference of two cubes.

Here is the formula for the difference of two cubes:

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

You should stop now to see if this is sufficient information for you to solve the limit.

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that first factor a - b is what we are going to say is (x+1)^(1/3) - 1, or in other words,
a = (x + 1)^(1/3)
b = 1

Is that enough?

So if we multiply the numerator and denominator by the second factor (a^2 + ab + b^2) that should not change anything. (we are multiplying by 1/1)

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this looks nasty

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

However the numerator is now [(x+1)^(1/3)]^3 - 1^3

That is where we got the formula from, remember.

The numerator then is (x+1) - 1 = x

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

The x's cancel, so we have

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

The threat of division by zero being removed, we can solve by substitution

Lim x->0 1/[(x+1)^(2/3) + (x+1)^(1/3)(1) + 1^2] =

1/[(0+1)^(2/3) + (0+1)^(1/3)(1) + 1^2] = 1/[1^(2/3) + 1^(1/3) + 1] = 1/3