# Thread: Need HELP with very tricky limit problem!

1. ## Need HELP with very tricky limit problem!

I need to find the limit using algebra of the limit as x approaches 1 of (x^(2/3)-1) / (x-1). I know from using L'Hospital's Rule that the answer is 2/3, however, I am not allowed to use that rule in the class, so does anybody know how to solve this with algebra. I would definitely appreciate the help.

2. Originally Posted by dahbch
I need to find the limit using algebra of the limit as x approaches 1 of (x^(2/3)-1) / (x-1). I know from using L'Hospital's Rule that the answer is 2/3, however, I am not allowed to use that rule in the class, so does anybody know how to solve this with algebra. I would definitely appreciate the help.
Hang on to your hats, folks. We're going to play!

Take a look at the numerator. This can be thought of as the difference between the square of two numbers, $x^{1/3}$ and 1. So
$x^{2/3} - 1 = (x^{1/3} - 1)(x^{1/3} + 1)$

In a similar fashion, the denominator is the difference between two cubes: $x^{1/3}$ and 1. The factorization here is $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ so
$x - 1 = (x^{1/3} - 1)(x^{2/3} + x^{1/3} + 1)$

Putting these together gives:
$\frac{x^{2/3} - 1}{x - 1} = \frac{(x^{1/3} - 1)(x^{1/3} + 1)}{(x^{1/3} - 1)(x^{2/3} + x^{1/3} + 1)}$ = $\frac{x^{1/3} + 1}{x^{2/3} + x^{1/3} + 1}$

And now you can take your limit.

-Dan

BTW Thank you for this problem. I haven't had this much fun doing a limit in ages!

3. Hello, topsquark!

Great insight . . . lovely and elegant!

Nice work, Dan!

4. Originally Posted by Soroban
Hello, topsquark!

Great insight . . . lovely and elegant!

Nice work, Dan!

(I'm blushing!)

-Dan