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
Looks good to me
Ok. Thanks. The answer key said -1 though. So I thought I might have done something wrong.
To solve the limit using L'Hospital's Rule, we begin by checking if it is an indeterminate form. In this case, it is indeed 0/0.
Next, we differentiate both the numerator and the denominator with respect to u.
The derivative of the numerator, (u - 1)^3, is 3(u - 1)^2.
The derivative of the denominator, (1/u) - u^2 + (3/u) - 3, can be a bit tricky. Let's simplify it first.
(1/u) - u^2 + (3/u) - 3 = 1/u^2 + 3/u - u^2 - 3
Now, differentiate each term:
d/du (1/u^2) = -2/u^3
d/du (3/u) = -3/u^2
d/du (-u^2) = -2u
d/du (-3) = 0
Combine all the derivatives:
-2/u^3 - 3/u^2 - 2u + 0 = -2/u^3 - 3/u^2 - 2u
Now, substitute u = 1 into the derivatives:
-2/1^3 - 3/1^2 - 2(1) = -2 - 3 - 2 = -7
So, the new limit becomes:
lim u --> 1 of [3(u - 1)^2 / (-2/u^3 - 3/u^2 - 2u)]
Now, we can evaluate the limit. Substituting u = 1, we get:
lim u --> 1 of [3(1 - 1)^2 / (-2/1^3 - 3/1^2 - 2(1))]
= lim u --> 1 of [0 / (-2 - 3 - 2)]
= 0 / (-7)
= 0
Hence, the solution to the limit using L'Hospital's Rule is 0.