# How do you evaluate the limit (sin(2x))/(2x^2+x) as x approaches 0?

Oct 7, 2016

${\lim}_{x \to 0} \sin \frac{2 x}{2 {x}^{2} + x} = 2$

#### Explanation:

Use the fact that:

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

Hence:

${\lim}_{x \to 0} \sin \frac{2 x}{2 {x}^{2} + x} = {\lim}_{x \to 0} \frac{\sin \left(2 x\right) \div \left(2 x\right)}{\left(2 {x}^{2} + x\right) \div \left(2 x\right)}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \sin \frac{2 x}{2 {x}^{2} + x}} = {\lim}_{x \to 0} \frac{\sin \left(2 x\right) \div \left(2 x\right)}{x + \frac{1}{2}}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \sin \frac{2 x}{2 {x}^{2} + x}} = \frac{1}{\frac{1}{2}}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \sin \frac{2 x}{2 {x}^{2} + x}} = 2$

Oct 7, 2016

$2$

#### Explanation:

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

so

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