Finding the limit of a trigonometric function

Question

I got it interms of e but couldn't find the limit of the power $$\displaystyle \lim_{x \to \frac{\pi}{4}} {\sin 2x}^{\tan^2 2x}$$

Answer 1

I do not know if you are allowed to use Taylor series; so forgive me if I am off-topic.

If I correctly read the problem, you are concerned by the behavior of $$A={\sin (2x)}^{\tan^2 (2x)}$$ So,$$\log(A)=\tan^2 (2x) \log\Big(\sin(2x)\Big)$$ Around $x=\frac{\pi}{4}$ we have $$\sin(2x)=1-2 \left(x-\frac{\pi }{4}\right)^2+\frac{2}{3} \left(x-\frac{\pi }{4}\right)^4+O\left(\left(x-\frac{\pi }{4}\right)^6\right)$$ Now, use the fact that, for small values of $y$, $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ and so $$\log\Big(1+\sin(2x)\Big)=-2 \left(x-\frac{\pi }{4}\right)^2-\frac{4}{3} \left(x-\frac{\pi }{4}\right)^4+O\left(\left(x-\frac{\pi }{4}\right)^6\right)$$ Now, and this is slightly more difficult, around $x=\frac{\pi}{4}$, an expansion of $\tan(2x)$ is given by $$\tan(2x)=-\frac{1}{2 \left(x-\frac{\pi }{4}\right)}+\frac{2}{3} \left(x-\frac{\pi }{4}\right)+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$ which then implies $$\tan^2(2x)=\frac{1}{4 \left(x-\frac{\pi }{4}\right)^2}-\frac{2}{3}+\frac{4}{15} \left(x-\frac{\pi }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$ So, replacing and expanding leads to $$\log(A)=-\frac{1}{2}+\left(x-\frac{\pi }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$ $$A=\frac{1}{\sqrt{e}}+\frac{\left(x-\frac{\pi }{4}\right)^2}{\sqrt{e}}+O\left(\left(x-\frac{\pi }{4}\right)^4\right)$$ If you plot the original function and this approximation on the same graph, you should be amazed to notice how close they are in the range $0 \leq x \leq \frac{\pi}{2}$.

Interesting or not, if you consider the function $$f(x)={\sin (2x)}^{\tan^2 (2x)}$$ after simplifications and factoring the first derivative write $$f'(x)=\tan (x) \sec ^2(x) {\sin (2x)}^{\tan^2 (2x)}\Big(\cos (2 x)+2 \log \big(\sin (2 x)\big)\Big)$$ and this cancels for $2x=\frac{\pi}{2}$; then the limit is finite.

Answer 2

As $x \to \pi/4$, the exponent goes to $+\infty$. So because $\pi/4<1$, you can do this problem without any sophisticated tricks: $x^{\tan^2(2x)} \to 0$ so the limit is $\sin(0)=0$.

Answer 3

Hint: $$\exp(\log((\sin 2x)^{\tan^2 2x})=\exp(\tan^2 2x\log \sin 2x).$$ As $x\to \pi/4$, $\tan 2x =\sin 2x/\cos 2x\sim 1/\cos 2x$. $$\lim_{x\to \pi/4} \cos2x = \lim_{y\to 0} \cos(\pi/2+2y) = \lim_{y\to 0} -\sin^22y .$$ $$\lim_{x\to \pi/4} \sin2x -1 = \lim_{y\to 0} \sin(\pi/2+2y) -1 = \lim_{y\to 0} \cos 2y-1 =\lim_{y\to 0} -\sin^2y.$$ Since $\log (1+y)\sim y$ for $y\to 1$, Then plug these approximations in the original expression inside $\exp$ to get $$\lim_{x\to \pi/2}\tan^2 2x\log \sin 2x= \lim_{y\to 0} \frac1{\sin^4 2y}(-\sin^2y)=-\infty,$$ hence your original limit is $0$.