Question

lim 1 - cos5x/1 - cos6x
x-0
pls solve!

Answer 1

Have a nice time.......

Answer 2

answer is 25/36

we have to find $\displaystyle\lim_{x\to 0}\frac{1-cos5x}{1-cos6x}$

we know, 1 - cos2α = 2sin²α

so, 1 - cos5x = 2sin²(5x/2)

1 - cos6x = 2sin²(3x)

now, $\displaystyle\lim_{x\to 0}\frac{1-cos5x}{1-cos6x}=\displaystyle\lim_{x\to p}\frac{2sin^2{5x/2}}{2sin^2(3x)}$

= $\displaystyle\lim_{x\to 0}\frac{sin^2(5x/2)}{sin^2(3x)}$

= $\displaystyle\lim_{x\to 0} \frac{\left(\frac{sin(5x/2)}{5x/2}\left)^2\times(5x/2)^2}{\left(\frac{sin(3x)}{3x}\right)^2\times(3x)^2}$

= $\displaystyle\lim_{x\to 0}\frac{25x^2}{4\times9x^2}$

= 25/36

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