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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\lim_{x \rightarrow \pi/4} (\sin 2x)^{\tan ^{2}(2x)}\] This is the problem
i'm not allowed to use l'hospital @OOOPS
we could try direct substitution, I guess

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

= sin (pi/2) ^ tan^2(pi/2)
what would we substitute in place of what??
x = pi/4, so 2x = pi/2
it gives me the form\[1^{\infty} \]
yes - i got confused the - I thought tan pi/2 = 1!!!
- a mental aberration!
no matter man:)
i think that there may be some misprint in the question
come on man i'm closing this question!
http://www.wolframalpha.com/input/?i=limit+as+x+approaches+pi%2F4+++of+[+sin%282x%29]^+%28tan2x%29
that's cool and that's the correct answer but i want it with a method:)
yea I know I guess you need to take logs - I tried that but could only get so far.
limits are not my strong point
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
That is lny =0 Hence y = e^0 =1 back to \(y = lim_{x\rightarrow \pi/4}(sin(2x)^{tan^2(2x)} =1 \)
I was waiting for closing because I violated the code when giving you the whole stuff!!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question