L' Hospital's Rule Calculus?

Question

need to apply l' hospitals rule to each and stat the indeterminate form for each. Thank you

lim x/sin(x)
x->0

lim (1-cos(x))/x
x->0

lim (4sin(2x))/(5sin(3x))
x->0

lim (4x^2 -3x+2)/(7x^2 +2x-1)
x->infinity

Answer 1

Take the derivative of top & bottom of the fractions, & put them in place of the original values, then set x value to the limit.

lim x/sin(x) = lim 1/cos(x) = 1 (all limits as x → 0)

lim (1-cos(x))/x = lim sin(x)/1 = 0 (all limits as x → 0)

lim (4 sin2x)/(5 sin3x) = lim [(4×2cos2x+0)/(5×3cos3x)] = 8/15 (all limits as x → 0)

If the rule gives a result that is still 0/0 or ∞/∞ in the limit, apply it again…

lim [(4x²−3x+2)/(7x²+2x−1)] = lim [(8x−3)/(14x+2)] = lim (8/14) = 8/14 (all limits as x → ∞ )

Answer 2

lim (x→0) x/sin x =
lim (x→0) 1/cos x =
1/1 =
1

lim (x→0) (1-cos x)/x =
lim (x→0) sin x / 1 =
0/1 =
0

lim (x→0) (4 sin 2x)/ 5 sin 3x) =
lim (x→0) (8 cos 2x) / (15 cos 3x) =
8/15

lim (x→∞) (4x² - 3x +2) / (7x² + 2x - 1) =
lim (x→∞) (4 - 3/x + 2/x²) / (7 + 2/x - 1/x²) =
(4 - 0 - 0) / (7 + 0 - 0) =
4/7