precalc help please..?

Question

lim as x approaches 0 of the function:

sin(x)/ [2x^(2-x)]

Answer 1

lim {sin(x)/ [2x^(2-x)]} =
x→0

lim{(1/2)[x^(x-2)]sin(x)} = 0
x→0

Answer 2

The expression can be broken up as follows:
(1/2) * [sin(x)/x] * [1/(x^(1-x))]
When x->0 the 1st term tends to 1/2, the 2nd term tends to 1, and the 3rd tends to infinity.
Therefore, the whole expression tends to infinity.