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

1. anonymous

2. anonymous

let me check it once more

3. anonymous

4. anonymous

what is "^@"

5. anonymous

I am sorry, the answer is 37 / 121, but the algebra is tiresome

6. anonymous

give me some time to post the full solution.

7. anonymous

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

8. anonymous

yep, L'Hopitals Rule twice ---> limit = 37/121

9. anonymous

yeah, that is much easier :)

10. anonymous

nice solution

11. anonymous

I will not forget L'Hospital anymore

12. anonymous

understood all part you had written. thank you very much for your help.

13. anonymous

do u have anything solution?

14. anonymous

I am sorry, but could not understand your question :), seemingly I dont have solution to this question

15. anonymous

((sin11x)^2)