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rajat97

  • one year ago

lim x->pi/4 {sin(2x)}^tan^2(2x) = ?

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  1. rajat97
    • one year ago
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    \[\lim_{x \rightarrow \pi/4} (\sin 2x)^{\tan ^{2}(2x)}\] This is the problem

  2. rajat97
    • one year ago
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    i'm not allowed to use l'hospital @OOOPS

  3. welshfella
    • one year ago
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    we could try direct substitution, I guess

  4. welshfella
    • one year ago
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    = sin (pi/2) ^ tan^2(pi/2)

  5. rajat97
    • one year ago
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    what would we substitute in place of what??

  6. welshfella
    • one year ago
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    x = pi/4, so 2x = pi/2

  7. rajat97
    • one year ago
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    it gives me the form\[1^{\infty} \]

  8. welshfella
    • one year ago
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    yes - i got confused the - I thought tan pi/2 = 1!!!

  9. welshfella
    • one year ago
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    - a mental aberration!

  10. rajat97
    • one year ago
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    no matter man:)

  11. rajat97
    • one year ago
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    i think that there may be some misprint in the question

  12. rajat97
    • one year ago
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    come on man i'm closing this question!

  13. welshfella
    • one year ago
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    http://www.wolframalpha.com/input/?i=limit+as+x+approaches+pi%2F4+++of+ [+sin%282x%29]^+%28tan2x%29

  14. rajat97
    • one year ago
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    that's cool and that's the correct answer but i want it with a method:)

  15. welshfella
    • one year ago
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    yea I know I guess you need to take logs - I tried that but could only get so far.

  16. welshfella
    • one year ago
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    limits are not my strong point

  17. anonymous
    • one year ago
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    Since you closed this question, I think I can work on it. :) let \(y = lim_{x\rightarrow \pi/4}(sin(2x)^{tan^2(2x)}\) \(lny = lim_{x\rightarrow \pi/4} tan^2(2x) ln sin(2x)\) Now, I work on the inside term only, will put back \(tan^2(2x) ln (sin (2x))\). I let ln(sin(2x)) aside also because as x approach (pi/4) , this guy =0. And I will use it later. So, I just have to calculate \(tan^2 (2x)\) \(tan^2(2x) = \dfrac{sin^2(2x)}{cos^2(2x)}=\dfrac{sin^2(2x)}{1-sin^2(2x)}\) multiply both sides by 4x^2, I have \(\dfrac{(4x^2 *sin(2x)*sin(2x))/4x^2 }{(4x^2(1-sin^2(2x))/4x^2}\) distribute to the terms I want only \(\dfrac{\cancel {4x^2}{\dfrac{sin(2x )}{2x}}*\dfrac{sin(2x)}{2x}}{\cancel{4x^2}*(\dfrac{1}{4x^2}-(\dfrac{sin(2x)}{2x}*\dfrac{sin(2x)}{2x})}\) as x approaches pi/4, this guy = \(\dfrac{1}{\dfrac{1}{\pi^2/4}-1}\) Hence combine with ln (sin(2x) which is approach 0 when x approaches pi/4 We have the whole limit =0

  18. anonymous
    • one year ago
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    That is lny =0 Hence y = e^0 =1 back to \(y = lim_{x\rightarrow \pi/4}(sin(2x)^{tan^2(2x)} =1 \)

  19. anonymous
    • one year ago
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    I was waiting for closing because I violated the code when giving you the whole stuff!!!

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