# Question b5c34

Sep 16, 2017

We seek:

${\lim}_{x \rightarrow 0} \frac{\tan 2 x - \sin 2 x}{x} ^ 2 = 0$

#### Explanation:

We seek:

$L = {\lim}_{x \rightarrow 0} \frac{\tan 2 x - \sin 2 x}{x} ^ 2$

If we put $x = 09$ we note both numerator and denominator are both $0$, so we have an indeterminate form $\frac{0}{0}$, and as such we can apply L'Hôpital's Rule to give:

$L = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} \left(\tan 2 x - \sin 2 x\right)}{\frac{d}{\mathrm{dx}} {x}^{2}}$
$\setminus \setminus \setminus = {\lim}_{x \rightarrow 0} \frac{2 {\sec}^{2} x - 2 \cos 2 x}{2 x}$
$\setminus \setminus \setminus = {\lim}_{x \rightarrow 0} \frac{{\sec}^{2} x - \cos 2 x}{x}$

Again, we have an indeterminate form $\frac{0}{0}$, and as such we can apply L'Hôpital's Rule again to give:

$L = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} \left({\sec}^{2} x - \cos 2 x\right)}{\frac{d}{\mathrm{dx}} x}$
$\setminus \setminus \setminus = {\lim}_{x \rightarrow 0} \frac{2 \sec x \sec x \tan x + 2 \sin x}{1}$
$\setminus \setminus \setminus = 2 {\lim}_{x \rightarrow 0} 2 \sec {x}^{2} \tan x + \sin x$
$\setminus \setminus \setminus = 2 \left(0 + 0\right)$
$\setminus \setminus \setminus = 0$

Sep 16, 2017

$0.$

#### Explanation:

Let us find the Limit L without using L'Hospital's Rule.

We will use ${\lim}_{\theta \to 0} \tan \frac{\theta}{\theta} = 1.$

Knowing that, tan 2x=(2tanx)/(1-tan^2x), &, sin2x=(2tanx)/(1+tan^2x),#

we have,

$\tan 2 x - \sin 2 x ,$

$= \frac{2 \tan x}{1 - {\tan}^{2} x} - \frac{2 \tan x}{1 + {\tan}^{2} x} ,$

$= 2 \tan x \left\{\frac{\left(1 + {\tan}^{2} x\right) - \left(1 - {\tan}^{2} x\right)}{\left(1 - {\tan}^{2} x\right) \left(1 + {\tan}^{2} x\right)}\right\} ,$

$= \frac{4 {\tan}^{3} x}{1 - {\tan}^{4} x} \times {x}^{3} / {x}^{3} ,$

$= 4 \cdot {\left(\tan \frac{x}{x}\right)}^{3} \cdot \frac{{x}^{2} \cdot x}{1 - {\tan}^{4} x} .$

$\Rightarrow \frac{\tan 2 x - \sin 2 x}{x} ^ 2 = 4 \cdot {\left(\tan \frac{x}{x}\right)}^{3} \cdot \frac{x}{1 - {\tan}^{4} x} .$

$\therefore L = {\lim}_{x \to 0} 4 \cdot {\left(\tan \frac{x}{x}\right)}^{3} \cdot \frac{x}{1 - {\tan}^{4} x} ,$

$= 4 \cdot {1}^{3} \cdot \frac{0}{1 - {0}^{4}} ,$

$\Rightarrow L = 0.$

Enjoy Maths.!