How to solve this limit \lim_{x\to pi/4} (sin2x)^(tan^2(2x))  ?

Mar 2, 2018

$\frac{1}{\sqrt{e}}$

Explanation:

The trick here is to use L'Hopital's Rule. although you might say there are no fractions but we can convert them into it. here is the start of my solution

And I want to clarify, I will be using the variable which you used as y and not x and use x as equal to 2y

${\lim}_{y \to \frac{\pi}{4}} {\left(\sin \left(2 y\right)\right)}^{{\tan}^{2} \left(2 y\right)}$

to simplify,
let $x$ be $2 y$
Realize that the equation is equal to

${\lim}_{x \to \frac{\pi}{2}} {\left(\sin \left(x\right)\right)}^{{\tan}^{2} \left(x\right)}$

as When $y$ approaches $\frac{\pi}{4}$, $x$ approaches $\frac{\pi}{2}$ as $x = 2 y$

Now, Break this up into an exponential with the base e

therefore,

lim _(x-> pi/2) e^(tan^2(x) * ln(sin(x))
using exponential rules

now, Bring the limit inside the equation using the Composition Law
https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.htmlComposition_Law

therefore the above equation is equal to
 e^( lim _(x-> pi/2) tan^2(x) * ln(sin(x))

Now Lets just focus on the Limit and try to use L'hôpital's Rule on it, but first, we need to convert it into a division of two functions

therefore,
${\lim}_{x \to \frac{\pi}{2}} {\tan}^{2} \left(x\right) \cdot \ln \left(\sin \left(x\right)\right)$
$=$
${\lim}_{x \to \frac{\pi}{2}} \ln \frac{\sin \left(x\right)}{\frac{1}{\tan} ^ 2 \left(x\right)}$

I assume you already know how to take the derivative of the two functions $\ln \left(\sin \left(x\right)\right)$ and $\frac{1}{\tan} ^ 2 \left(x\right)$ which comes out as

(d/dx ln(sin(x))) / (d/dx 1/tan^2(x)

$= \cot \frac{x}{- 2 \cot \left(x\right) \cdot {\csc}^{2} \left(x\right)}$

after cancelling cot (x), we Get
$= \frac{1}{- 2 \cdot {\csc}^{2} \left(x\right)}$

therefore the limit is
${\lim}_{x \to \frac{\pi}{2}} \frac{1}{- 2 \cdot {\csc}^{2} \left(x\right)}$

which is continuous and defined at $x = \frac{\pi}{2}$ and ${\csc}^{2} \left(\frac{\pi}{2}\right)$ is 1, therefore the entire power becomes $\frac{1}{-} 2 = - \frac{1}{2}$

and don't forget, this is the power $e$ is raised to so the Complete limit is ${e}^{- \frac{1}{2}}$
$=$
$\frac{1}{\sqrt{e}}$

Mar 2, 2018

$\frac{1}{\sqrt{e}}$

Explanation:

Making $y = \frac{\pi}{4} - x$

sin(2x)^(tan^2(2x))=sin(pi/2-2y)^(tan^2(pi/2-2y)) = cos(2y)^(cot^2(2y)#

now $x \to \frac{\pi}{4} \Rightarrow y \to 0$ and then making $z = 2 y$

${\lim}_{x \to \frac{\pi}{4}} \sin {\left(2 x\right)}^{{\tan}^{2} \left(2 x\right)} \equiv {\lim}_{z \to 0} {\left(\cos z\right)}^{{\cot}^{2} z}$

but ${\cot}^{2} z = \frac{1}{\sin} ^ 2 z - 1 = \frac{1}{\frac{1}{2} \left(1 - \cos \left(2 z\right)\right)} - 1$ then

${\left(\cos z\right)}^{{\cot}^{2} z} = {\left(\cos z\right)}^{\frac{2}{1 - \cos \left(2 z\right)}} / \cos z$

but $\frac{2}{1 - \cos \left(2 z\right)} = \frac{1}{1 - {\cos}^{2} z} = \frac{1}{2} \left(\frac{1}{1 - \cos z} - \frac{1}{1 + \cos z}\right)$

then

${\lim}_{z \to 0} {\left(\cos z\right)}^{{\cot}^{2} z} = {\lim}_{z \to 0} {\left(\cos z\right)}^{\frac{1}{2} \left(\frac{1}{1 - \cos z} - \frac{1}{1 + \cos z}\right)} / \cos z =$

${\lim}_{z \to 0} {\left(\cos z\right)}^{\frac{1}{2} \left(\frac{1}{1 - \cos z}\right)} / {\left(\cos z\right)}^{\frac{1}{2} \left(\frac{1}{1 + \cos z}\right)} \frac{1}{\cos} z =$

$= {\lim}_{z \to 0} {\left(\cos z\right)}^{\frac{1}{2} \left(\frac{1}{1 - \cos z}\right)}$

because

${\lim}_{z \to 0} \frac{1}{\cos z {\left(\cos z\right)}^{\frac{1}{2} \left(\frac{1}{1 + \cos z}\right)}} = 1$

Now expanding $\cos z = 1 - \frac{1}{2} {z}^{2} + O \left({z}^{4}\right)$ near the origin we have

${\lim}_{z \to 0} {\left(\cos z\right)}^{\frac{1}{2} \left(\frac{1}{1 - \cos z}\right)} \equiv {\lim}_{z \to 0} {\left(1 - {z}^{2} / 2 + O \left({z}^{4}\right)\right)}^{\frac{1}{2} \left(\frac{1}{{z}^{2} / 2 - O \left({z}^{4}\right)}\right)} = \sqrt{{\lim}_{z \to 0} {\left(1 - {z}^{2} / 2 + O \left({z}^{4}\right)\right)}^{\frac{1}{{z}^{2} / 2 - O \left({z}^{4}\right)}}} = \frac{1}{\sqrt{e}}$