Real Analysis limit problem

[kedlav]

Having a bit of trouble with an extra credit limit problem. The problem does NOT allow the use of L'Hopital's Rules(grr).

lim [x^(7/3) - 1]/[x^(1/2) - 1]
x->1

or

Obviously, begin by multiplying by the reciprocal of the denominator, hence it is equal to

lim
x->1 [x^(17/6) + x^(7/3) - x^(1/2) - 1]/[x-1]

Here's the issue. Using long division for polynomials, I keep getting what appears to be an infinite series that looks like x^(11/6) + x^(8/6) + x^(5/6) + x^(2/6) + x^(-1/6) - x^(-3/6) +x^(-4/6) + x^(-7/6) - x^(-9/6) + x^(-10/6) + x^(-13/6) - x^(-15/6) + [x^(-10/6) + x^(-13/6) + x^(-15/6) - 1]/[x-1]. Any help would be much appreciated.

Edit: I've also ran through trying to use the 'Squeeze Theorem' and a few other tricks, yet still remain stumped.[/img]

 

[tkhunny]

Have you noticed that your numerator can be usefully factored?

\(\displaystyle x^{7/3}-1\,=\,(x^{1/3}-1)(x^{2}+x^{5/3}+x^{4/3}+x+x^{2/3}+x^{1/3}+1)\)

Then, you may wish to notice that \(\displaystyle (x^{1/3}-1)*(x^{2/3}+x^{1/3}+1)\,=\,(x-1)\)

Finally, \(\displaystyle (x-1)\,=\,(x^{1/2}+1)(x^{1/2}-1)\)

That wasn't much fun.