M2 Limits

1.
Since cos 2x = 1 - 2 sin^2 x
so cos 3x = cos 2(3x/2) = 1 - 2 sin^2(3x/2)
so 1 - cos 3x = 2 sin^2 (3x/2)
(1 - cos 3x)/x^2 = [2 sin ^2(3x/2)]/x^2 = 2 [sin (3x/2)/x]^2
= 2 [ (3/2) sin (3x/2)/(3x/2)]^2
Since limit sin x/x = 1 when x tends to zero
So answer = 2 (3/2)^2 = 9/4.
2.
(cos x - cos 3x)/x^2 = (2 sin x sin 2x)/x^2 = 2 (sin x/x)[2 sin 2x/(2x)]
So limit when x tends to zero = 2(1)(2) = 4.
3. (1 + 1/4n)^n = (1 + 1/4n)^(4n/4) = [(1 + 1/4n)^(4n)]^(1/4)
so limit when n tends infinity = e^(1/4)
For (1 + 1/4n)^1, limit when n tends to infinity = 1
so limit of (1 + 1/4n)^(n + 1) = e^(1/4)(1) = e^(1/4).

2012-11-25 21:19:25 補充：
Correction : Answer for Q1 should be 9/2. Sorry.

Note: cos(0) = 1, sin(0) = 0

1.
lim (x -> 0) (1 - cos3x) / x^2 　　[0 / 0 form -> L'Hopital's rule]
= lim (x -> 0) 3sin3x / 2x 　　[0 / 0 form -> L'Hopital's rule]
= lim (x -> 0) 9cos3x / 2
= 9(1)/2
= 9/2

2.
lim (x -> 0) (cosx - cos3x) / x^2　　[0 / 0 form -> L'Hopital's rule]
= lim (x -> 0) [-sinx + 3sin3x] / 2x　　[0 / 0 form -> L'Hopital's rule]
= lim (x -> 0) [-cosx + 9cos3x] / 2
= (-1 + 9)/2
= 4

3.
lim (n -> ∞) (1 + 1/4n)^(n + 1)

Let
y = (1 + 1/4n)^(n + 1)
i.e. ln y = (n + 1) ln(1 + 1/4n)

lim (n -> ∞) ln y
= lim (n -> ∞) (n + 1) ln(1 + 1/4n)
= lim (n -> ∞) ln(1 + 1/4n) / [1 / (n + 1)]　　[0 / 0 form -> L'Hopital's rule]
= lim (n -> ∞) [-1/(4n^2 + n)] / [-1/(n + 1)^2]
= lim (n -> ∞) (n^2 + 2n + 1) / (4n^2 + n)
= lim (n -> ∞) (1 + 2/n + 1/n^2) / (4 + 1/n)
= (1 + 0 + 0) / (4 + 0)
= 1/4

->
lim (n -> ∞) y = e^(1/4)

i.e. lim (n -> ∞) (1 + 1/4n)^(n + 1) = e^(1/4)