# Question #f6818

Jul 21, 2016

$\frac{1}{4}$

#### Explanation:

${\lim}_{x \to 0} \frac{2 x - \sin x}{3 x + \sin x}$

indeterminate $\frac{0}{0}$, so applying L'Hopital's rule

$= {\lim}_{x \to 0} \frac{2 - \cos x}{3 + \cos x}$

cos x is continuous through the limit, and the expression overall is continuous, so we can apply x = 0 to the limit

$= \frac{2 - 1}{3 + 1} = \frac{1}{4}$

Jul 21, 2016

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

#### Explanation:

$\frac{2 x - \sin x}{3 x + \sin x}$.
We want to get the expression $\sin \frac{x}{x}$. Since the limit at $0$ doesn't care what happens when $x = 0$ we can multiply by $1$ in the form $\frac{\frac{1}{x}}{\frac{1}{x}}$.

We get

$\frac{2 x - \sin x}{3 x + \sin x} = \left(\frac{\frac{1}{x}}{\frac{1}{x}}\right) \cdot \frac{\left(2 x - \sin x\right)}{\left(3 x + \sin x\right)}$

$= \frac{2 - \sin \frac{x}{x}}{3 + \sin \frac{x}{x}}$

${\lim}_{x \rightarrow 0} \frac{2 x - \sin x}{3 x + \sin x} = {\lim}_{x \rightarrow 0} \frac{2 - \sin \frac{x}{x}}{3 + \sin \frac{x}{x}}$

$= \frac{2 - \left(1\right)}{3 + \left(1\right)}$

$= \frac{1}{4}$