find the limit

Question

lim (sin^2 * 2x)/x^2

x→0

 

 

Answer 1

Find the limit sin^2(2x)/x^2, as x -> 0.

sin^2(2x) = (1/2)(1 - cos(4x))

Macalurin series for cos(t) is,

cos(t) = 1 - t^2/2! + t^4/4! - t^6/6! - ...

So for cos(4x), we will have,

cos(4x) = 1 - (4x)^2/2! + (4x)^4/4! - (4x)^6/6! - ...

cos(4x) = 1 - 16x^2/2! + 256x^4/4! - 4096x^6/6! - ...

cos(4x) = 1 - 8x^2 + (32/3)x^4 - (256/45)x^6 - ...

Then (1 - cos(4x)) is,

(1 - cos(4x)) = 8x^2 - (32/3)x^4 + (256/45)x^6 - ...

And sin^2(2x)/x^2 = (1/2)(1 - cos(4x))/x^2 = (4x^2 - (16/3)x^4 + (128/45)x^6 - ...)/x^2

sin^2(2x)/x^2 = 4 - (16/3)x^2 + (128/45)x^4 - ...)

So limit, as x -> 0 = 4

Answer: limit =4