Question

Lim x tends to 1-cos 4 theta /1-cos 6 theta

Answer 1

Please see the attachment

Answer 2

answer is 4/9

we have to find $\displaystyle\lim_{x\to 0}\frac{1-cos4\theta}{1-cos6\theta}$

we know,

1 - cos2α = 2sin²α

so, 1 - cos4x = 1 - cos2(2x) = 2sin²(2x)

1 - cos6x = 1 - cos2(3x) = 2sin²(3x)

$\displaystyle\lim_{x\to 0}\frac{2sin^2(2x)}{2sin^2(3x)}$

= $\displaystyle\lim_{x\to 0}\frac{sin^2(2x)}{sin^2(3x)}$

= $\frac{\displaystyle\lim_{x\to 0}\frac{sin^2(2x)}{(2x)^2}\times 4x^2}{\displaystyle\lim_{x\to 0}\frac{sin^2(3x)}{(3x)^2}\times 9x^2}$

= 4x²/9x²

= 4/9

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