Question

Please solve this question:

Lim. (1 - cos 2x) / (cos 2x - cos 8x)

x-> 0

Answer 1

Answer:

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Answer 2

The solution for Lim. (1 - cos 2x) / (cos 2x - cos 8x)

x-> 0 is 1/7.

   

Step-by-step explanation:

Given:

$\lim_{x \to 0} \frac{1 - cos 2x}{cos 2x - cos 8x}$

=$\lim_{x \to 0} \frac{1 - cos 2x}{cos 2x -{2(2cos^22x-1)}-1}$

=$\lim_{x \to \0} \frac{1-cos(2x)}{cos 2x-4cos^22x+3}$

=$\lim_{x \to 0}\frac{1-cos(2x)}{-4 cos2^2x + cos 2x +3}$

=$\lim_{x \to 0}\frac{1-cos(2x)}{-[4 cos2^2x - cos 2x -3]}$

=$\lim_{x \to \0} \frac{-(cos 2x - 1)}{-[4 cos22x - cos 2x - 3}$

Splitting the middle term of the denominator

=$\lim_{x \to \0}\frac{(cos 2x - 1)}{4 cos^22x - 4 cos 2x + 3 cos 2x - 3}$

=$\lim_{x \to \0}\frac{(cos 2x - 1)}{4 cos 2x(cos 2x - 1) + 3(cos 2x - 1)}$

=$\lim_{x \to \0}\frac{(cos 2x - 1)}{(cos 2x - 1)(4 cos 2x + 3)}$

=$\lim_{x \to \0}\frac{1}{(4 cos 2x + 3)}$

=$\frac{1}{4 cos (0) + 3}$

=$\frac{1}{1\times4+3}$    [cos(0)=1]

=$\frac{1}{7}$