3) lim[x->0] [(x^2-2x-1)/(x^2+3x-1)]^[(x^2+2)/x] ?

Oh snap that's some serious $h!t right there.

Ok, so the x^2 terms in the base AND exponent can be neglected. So it becomes this:
((-2x-1)/(3x-1))^(2/x)
= (1-5x/(3x-1))^(2/x)
And I can neglect the 3x in the denominator of the base for it being near 0:
= (1+5x)^(2/x)
And this is e to the 10th power. exp(10). That's about 22026.466.

By the way, you want to put in some spaces between things so yahoo doesn't turn it into ....
lim[x->0] [(x^2-2x-1)/(x^2+ 3x-1)]^[(x^2+2)/x] see how I put in the space before the 3x? If there are too many characters in a row without a space yahoo stops and shows ... though it shows the full question in the list of questions so I could see what it should have said.

Let L = lim(x→0) [(x²-2x-1)/(x²+3x-1)]^[(x²+2)/x].

Take natural logs of both sides:
ln L = lim(x→0) ((x²+2)/x) * ln[(x²-2x-1)/(x²+3x-1)]
.......= lim(x→0) ln[(x²-2x-1)/(x²+3x-1)] / [x/(x²+2)]

Since this is of the form 0/0, we can apply L'Hopital's Rule.
ln L = lim(x→0) (d/dx) ln[(x²-2x-1)/(x²+3x-1)] / (d/dx) [x/(x²+2)]
.......= lim(x→0) (d/dx) [ln(x²-2x-1) - ln(x²+3x-1)] / (d/dx) [x/(x²+2)]
.......= lim(x→0) [(2x-2)/(x²-2x-1) - (2x+3)/(x²+3x-1)] / [(2-x²)/(x²+2)²]
.......= [(-2)/(-1) - 3/(-1)] / (2/4)
.......= 10.

Hence, L = e¹⁰.

Double checked on Wolfram Alpha:
https://www.wolframalpha.com/input/?i=lim(x%E2%86%920)+%5B(x%C2%B2-2x-1)%2F(x%C2%B2%2B3x-1)%5D%5E%5B(x%C2%B2%2B2)%2Fx%5D

I hope this helps!