# lim_(x->pi/2)(sinx)^(1/sin(2x)) = ?

Jan 27, 2018

undefined

#### Explanation:

Solve by direct substitution

lim_(x->pi/2)(sin x)^(1/(sin 2x)) = (sin (pi/2))^(1/(sin (pi))

$= {0}^{\frac{1}{0}} = \infty \mathmr{and} \text{undefined}$

Apply L'Hopital's rule but to do this we have to modify the function

$y \left(x\right) = {\left(\sin x\right)}^{\frac{1}{\sin 2 x}}$

Take natural logarithm of both sides

$\ln \left(y\right) = \ln {\left(\sin x\right)}^{\frac{1}{\sin 2 x}} = \frac{1}{\sin \left(x\right)} \ln \left(\sin \left(2 x\right)\right)$

$\ln \left(y\right) = \ln \frac{\sin \left(x\right)}{\sin \left(x\right)}$

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

$= \text{undefined}$

Now we can apply L'Hopital's rule

Numerator

$\frac{d}{\mathrm{dx}} \sin \left(x\right) = \cos \left(x\right)$

Denominator

$\frac{d}{\mathrm{dx}} \left(\sin \left(2 x\right)\right) = \frac{d}{\mathrm{du}} \left(\sin \left(u\right)\right) \frac{d}{\mathrm{dx}} \left(2 x\right)$

$= \cos \left(u\right) \cdot 2$

$= \cos \left(2 x\right) \cdot 2$

Divide numerator by denominator

$\cos \frac{x}{\cos \left(2 x\right) \cdot 2} = \cos \frac{\frac{\pi}{2}}{\cos \left(\pi\right) \cdot 2} = \frac{0}{0}$

So again we derivate numerator and denominator

Numerator

$\frac{d}{\mathrm{dx}} \cos \left(x\right) = - \sin \left(x\right) = - 1$

Denominator

$\frac{d}{\mathrm{dx}} \left(\cos \left(2 x\right) \cdot 2\right) = 2 \frac{d}{\mathrm{du}} \cos \left(u\right) \frac{d}{\mathrm{dx}} \left(2 x\right)$

$= - 4 \sin \left(2 x\right)$

$= 0$

I used a calculator to find where the function becomes defined but you no matter how much you derivate the answer remain undefined

Numerator

$\frac{d}{\mathrm{dx}} \left(- \sin \left(x\right)\right) \rightarrow - \cos \left(x\right) \rightarrow \frac{d}{\mathrm{dx}} \left(- \cos \left(x\right)\right) \rightarrow \sin \left(x\right) \rightarrow \sin \left(\frac{\pi}{2}\right) = 1$

Denominator

$\frac{d}{\mathrm{dx}} \left(- 4 \sin \left(2 x\right)\right) \rightarrow - 8 \setminus \cos \left(2 x \setminus\right) \rightarrow \frac{d}{\mathrm{dx}} \left(- 8 \setminus \cos \left(2 x \setminus\right)\right) \rightarrow 16 \setminus \sin \setminus \left(2 x \setminus\right) \rightarrow 16 \setminus \sin \setminus \left(\pi \setminus\right) = 0$

Jan 27, 2018

${\lim}_{x \to \frac{\pi}{2}} {\left(\sin x\right)}^{\frac{1}{\sin} \left(2 x\right)} = 1$

#### Explanation:

Near $x = \frac{\pi}{2}$ we have

$\sin x = 1 - \frac{1}{2} {\left(x - \frac{\pi}{2}\right)}^{2} + \frac{1}{24} {\left(x - \frac{\pi}{2}\right)}^{2} + O \left({\left(x - \frac{\pi}{2}\right)}^{5}\right)$

and

$\sin \left(2 x\right) = - 2 \left(x - \frac{\pi}{2}\right) + \frac{4}{3} {\left(x - \frac{\pi}{2}\right)}^{3} + O \left({\left(x - \frac{\pi}{2}\right)}^{5}\right)$

Then making $x - \frac{\pi}{2} = y$

${\lim}_{x \to \frac{\pi}{2}} {\left(\sin x\right)}^{\frac{1}{\sin} \left(2 x\right)} \equiv {\lim}_{y \to 0} {\left(1 - \frac{1}{2} {y}^{2} + O \left({y}^{4}\right)\right)}^{\frac{1}{- 2 y + O \left({y}^{3}\right)}}$

$= {\lim}_{y \to 0} {1}^{- \frac{1}{2 y}} = 1$

${\lim}_{x \to \frac{\pi}{2}} {\left(\sin x\right)}^{\frac{1}{\sin} \left(2 x\right)} = 1$