# Tanx/sin2x? limit x->0

Jun 22, 2018

${\lim}_{x \to 0} \tan \frac{x}{\sin} \left(2 x\right) = \frac{1}{2}$

#### Explanation:

Consider the fundamental trigonometric limit:

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

and note that also:

${\lim}_{x \to 0} \tan \frac{x}{x} = {\lim}_{x \to 0} \frac{1}{\cos} x \sin \frac{x}{x} = 1$

Then:

${\lim}_{x \to 0} \tan \frac{x}{\sin} \left(2 x\right) = {\lim}_{x \to 0} \frac{1}{2} \tan \frac{x}{x} \frac{2 x}{\sin} \left(2 x\right)$

${\lim}_{x \to 0} \tan \frac{x}{\sin} \left(2 x\right) = \frac{1}{2} {\lim}_{x \to 0} \tan \frac{x}{x} {\lim}_{x \to 0} \frac{2 x}{\sin} \left(2 x\right)$

${\lim}_{x \to 0} \tan \frac{x}{\sin} \left(2 x\right) = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$

Jun 22, 2018

${\lim}_{x \to 0} \tan \frac{x}{\sin 2 x} = \frac{1}{2}$

#### Explanation:

Let $L = {\lim}_{x \to 0} \tan \frac{x}{\sin 2 x}$.

$L = {\lim}_{x \to 0} \frac{\textcolor{red}{\sin \frac{x}{\cos} x}}{\textcolor{b l u e}{2 \sin x \cos x}}$

$L = {\lim}_{x \to 0} \frac{1}{2 {\cos}^{2} x} = \frac{1}{2 {\cos}^{2} 0} = \frac{1}{2}$