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anonymous
 4 years ago
lim as x  > 0 (1cos5xcos7x)/((sin11x)^@)
anonymous
 4 years ago
lim as x  > 0 (1cos5xcos7x)/((sin11x)^@)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me check it once more

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i want to know how did u find out the answer???please!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am sorry, the answer is 37 / 121, but the algebra is tiresome

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0give me some time to post the full solution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since, 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

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1yep, L'Hopitals Rule twice > limit = 37/121

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, that is much easier :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will not forget L'Hospital anymore

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0understood all part you had written. thank you very much for your help.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do u have anything solution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am sorry, but could not understand your question :), seemingly I dont have solution to this question
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