How do you solve this question without using L'Hopital's Rule?

The answer to a) is -∞.
The numerator approaches 2.296875 as x -> 0.5 from
the left. (Just plug in 0.5 and calculate.)
The denominator approaches 0 and is always
negative, so the answer is -∞.

b). Factor out an x to get
lim x -> 0 x/sin x * (2x+1)
The limit as x -> 0 of x/sin x = 1, so the answer is 1.

c). The trick here is to multiply numerator and denominator by
1 + cos 4x.
This gives
lim x-> 0 (1-cos 4x)(1+ cos 4x)/ [9x^ 2(1+ cos 4x)].
But the numerator is sin^ 2 4x and lim x-> 0 1 + cos 4x = 1
so we have to find
lim x->0 (sin 4x/3x)^ 2.
Now let u = 4x. Then 3x = 3u/4
So the problem becomes
lim u->0 (4/3 sin u/u)^ 2 = 16/9.
Hope that helps!

Actually for c the answer is 8/9, since 1+cos4x = 2.