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rajat97
 one year ago
lim x>pi/4 {sin(2x)}^tan^2(2x) = ?
rajat97
 one year ago
lim x>pi/4 {sin(2x)}^tan^2(2x) = ?

This Question is Closed

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \pi/4} (\sin 2x)^{\tan ^{2}(2x)}\] This is the problem

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0i'm not allowed to use l'hospital @OOOPS

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0we could try direct substitution, I guess

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0= sin (pi/2) ^ tan^2(pi/2)

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0what would we substitute in place of what??

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0x = pi/4, so 2x = pi/2

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0it gives me the form\[1^{\infty} \]

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0yes  i got confused the  I thought tan pi/2 = 1!!!

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0 a mental aberration!

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0i think that there may be some misprint in the question

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0come on man i'm closing this question!

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=limit+as+x+approaches+pi%2F4+++of+ [+sin%282x%29]^+%28tan2x%29

rajat97
 one year ago
Best ResponseYou've already chosen the best response.0that's cool and that's the correct answer but i want it with a method:)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0yea I know I guess you need to take logs  I tried that but could only get so far.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0limits are not my strong point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since 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 \(tan^2 (2x)\) \(tan^2(2x) = \dfrac{sin^2(2x)}{cos^2(2x)}=\dfrac{sin^2(2x)}{1sin^2(2x)}\) multiply both sides by 4x^2, I have \(\dfrac{(4x^2 *sin(2x)*sin(2x))/4x^2 }{(4x^2(1sin^2(2x))/4x^2}\) 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is lny =0 Hence y = e^0 =1 back to \(y = lim_{x\rightarrow \pi/4}(sin(2x)^{tan^2(2x)} =1 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was waiting for closing because I violated the code when giving you the whole stuff!!!
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