Question

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

Answer 1

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

Answer 2

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

Answer 3

we could try direct substitution, I guess

Answer 4

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

Answer 5

what would we substitute in place of what??

Answer 6

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

Answer 7

it gives me the form\[1^{\infty} \]

Answer 8

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

Answer 9

- a mental aberration!

Answer 10

no matter man:)

Answer 11

i think that there may be some misprint in the question

Answer 12

come on man i'm closing this question!

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

limits are not my strong point

Answer 17

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

Answer 18

That is lny =0
Hence y = e^0 =1
back to \(y = lim_{x\rightarrow \pi/4}(sin(2x)^{tan^2(2x)} =1 \)

Answer 19

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