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

1. rajat97

$\lim_{x \rightarrow \pi/4} (\sin 2x)^{\tan ^{2}(2x)}$ This is the problem

2. rajat97

i'm not allowed to use l'hospital @OOOPS

3. welshfella

we could try direct substitution, I guess

4. welshfella

= sin (pi/2) ^ tan^2(pi/2)

5. rajat97

what would we substitute in place of what??

6. welshfella

x = pi/4, so 2x = pi/2

7. rajat97

it gives me the form$1^{\infty}$

8. welshfella

yes - i got confused the - I thought tan pi/2 = 1!!!

9. welshfella

- a mental aberration!

10. rajat97

no matter man:)

11. rajat97

i think that there may be some misprint in the question

12. rajat97

come on man i'm closing this question!

13. welshfella

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

14. rajat97

that's cool and that's the correct answer but i want it with a method:)

15. welshfella

yea I know I guess you need to take logs - I tried that but could only get so far.

16. welshfella

limits are not my strong point

17. anonymous

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

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

I was waiting for closing because I violated the code when giving you the whole stuff!!!