lim as x - -> 0 (1-cos5xcos7x)/((sin11x)^@)

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lim as x - -> 0 (1-cos5xcos7x)/((sin11x)^@)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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answer is 0
let me check it once more
i want to know how did u find out the answer???please!

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Other answers:

what is "^@"
I am sorry, the answer is 37 / 121, but the algebra is tiresome
give me some time to post the full solution.
Since, Cos C + Cos D = 2.cos ((C + D) / 2)cos ((C -D) / 2) \[1 - \cos5x.\cos7x = 1 - (1/2) (\cos12x + \cos 2x)\] = 1/2(1 - COS12X) + 1/2 (1 - COS2X) = (1/2).2(SIN6X)^2 + (1/2).2(SINX)^2 = (SIN6X) ^ 2 + (SINX)^2 Hence the equation becomes -\[\sin ^{2}6x / \sin ^{2}11x + \sin ^{2}x/\sin ^{2}11x\] The first part \[= [(\sin ^{2}6x / (6x)^{2}) / (\sin ^{2}11x / (11x) ^{2})] (36 /121)\] Did you understand this part? (36 / 121 is introduced to nullify (6x)^2 and (11x)^2) Hence the first part becomes 36 / 121 Similarly the second part becomes 1 / 121 Hence it becomes in total = 37 / 121
yep, L'Hopitals Rule twice ---> limit = 37/121
yeah, that is much easier :)
nice solution
I will not forget L'Hospital anymore
understood all part you had written. thank you very much for your help.
do u have anything solution?
I am sorry, but could not understand your question :), seemingly I dont have solution to this question
((sin11x)^2)

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