Use the fact that limx→0 sin(x)/x = 1 to compute the following limits.

Question

(A.) lim x→0 sin(5x)/2x =

(b.) lim t→0 sin2(3t)/t =

(c) lim θ→0 sin(3θ)/tan(θ) =

Answer 1

lim_x→0 [sin(5x) / (2x)] = (5/2)lim_x→0[sin(5x) /(5x)] = (5/2)(1) = 5/2

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lim_x→0 [sin²(3x) / x] = 3lim_x→0 [sin(3x) / (3x)] lim_x→0(sin(3x)) = 3(1)(0) = 0

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lim_x→0 [sin(3x) / tanx] = lim_x→0 [sin(3x) cosx / sinx] =

lim_x→0 [(sin(3x)/(3x))(cosx)((3x)/(sinx)] = 3lim_x→0[sin(3x)/(3x)] lim_x→0(cosx) lim_x→0 [x / sinx]

= 3(1)(1)(1) = 3