What is wrong with this students work? (L'hospitals')?

Question

Lim (x^2-2x/x^2-sin x) = Lim (2x-2/2x-cosx) = Lim (2/2+sinx) = 1
x-> 0 x->0 x->0

Answer 1

The first thing to determine whether we can apply L'Hospital's rule is if it is in the form [0/0]

Let's take the first limit.

Lim( (x^2-2x)/(x^2 - sinx) )dx
x -> 0
This in the form (0 - 0)/(0 - 0) = 0/0, so we can freely use L'Hospital's rule.

Lim ( (2x - 2)/(2x - cos(x))dx
x -> 0
If we plug in x=0, we get the form [-2]/[0 - cos(0)], or
[-2]/(-cos(0)) = (-2)/(-1) = 2

At this point, we actually have a value for our limit, so we CANNOT use L'Hospital's rule (as it isn't in the form 0/0 anymore).

Answer 2

you take the LH because if you plug in 0 in the first one you get the denominator 0 and that can't happen. So after the first LH you can just plug 0 in:
Lim x->0 (2x-2/2x-cosx) = (2(0)-2)/(2(0)-cos(0)) = -2/-1 = 2

Answer 3

The problem was done after the first application of L'H.

Lim (2x-2)
--------------
2x - cos(x)

as x goes to 0 becomes -2 over -1 because lim of cos(x) is 1 as x goes to zero