Mar 18, 2017

$= \frac{1}{8}$

#### Explanation:

We can use the first few terms of the various Taylor Series here:

sin theta = theta - theta^3/(3!) + O(theta^5)

Differentiate that to get the $\cos$ series:

cos theta = 1 - theta^2/(2!) + O(theta^4)

Replace $\cot 4 x$ as $\frac{\cos 4 x}{\sin 4 x}$, set $\theta = 2 x$ and $\theta = 4 x$ as appropriate in the series expansions, and then insert the first few terms of those series to get:

lim_(x to 0) (x^2 (1 - (4x)^2/(2!)))/(((4x) - (4x)^3/(3!))((2x) - (2x)^3/(3!))

$= {\lim}_{x \to 0} \frac{{x}^{2} + O \left({x}^{4}\right)}{8 {x}^{2} + O \left({x}^{4}\right)}$

$= {\lim}_{x \to 0} \frac{1 + O \left({x}^{2}\right)}{8 + O \left({x}^{2}\right)} = \frac{1}{8}$

Mar 18, 2017

${\lim}_{x \to 0} \frac{{x}^{2} \cot 4 x}{\sin 2 x} = \frac{1}{8}$

#### Explanation:

Start by simplifying the expression considering that:

$\cot 4 x = \frac{\cos 4 x}{\sin 4 x} = \frac{{\cos}^{2} 2 x - {\sin}^{2} 2 x}{2 \sin 2 x \cos 2 x}$

so:

$\frac{\cot 4 x}{\sin 2 x} = \frac{{\cos}^{2} 2 x - {\sin}^{2} 2 x}{2 {\sin}^{2} 2 x \cos 2 x} = \frac{\cos 2 x}{2 {\sin}^{2} 2 x} - \frac{1}{2 \cos 2 x}$

Now we have:

${\lim}_{x \to 0} \frac{{x}^{2} \cot 4 x}{\sin 2 x} = {\lim}_{x \to 0} \frac{{x}^{2} \cos 2 x}{2 {\sin}^{2} 2 x} - {x}^{2} / \left(2 \cos 2 x\right)$

The second term is continuous in $x = 0$:

${\lim}_{x \to 0} {x}^{2} / \left(2 \cos 2 x\right) = {0}^{2} / 2 = 0$

so:

lim_(x->0) (x^2 cot4x)/(sin2x) = lim_(x->0) (x^2 cos2x)/(2sin^2 2 x) = lim_(x->0) ( cos2x)/(8((sin 2x)/(2x))^2

and using the limit:

${\lim}_{\theta \to 0} \sin \frac{\theta}{\theta} = 1$

we can conclude:

${\lim}_{x \to 0} \frac{{x}^{2} \cot 4 x}{\sin 2 x} = \frac{1}{8}$