Asked by Mary
Use L'Hospital's Rule to solve:
lim u --> 1 of (u-1)^3/ ((1/u) - u^2 + (3/u) - 3)
Ok, so what I thought was that it is type 0/0
so taking the derivatives of the top and bottom
3(u-1)^2 /(-u^-2 -2u - 3u^-3)
and subbing in u = 1
= 0/-6
= 0
lim u --> 1 of (u-1)^3/ ((1/u) - u^2 + (3/u) - 3)
Ok, so what I thought was that it is type 0/0
so taking the derivatives of the top and bottom
3(u-1)^2 /(-u^-2 -2u - 3u^-3)
and subbing in u = 1
= 0/-6
= 0
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