# How do you find the limit of  ( x + sin(x) ) / (x)  as x approaches 0?

May 22, 2018

$2$.

#### Explanation:

Knowing that, ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$, we have,

${\lim}_{x \to 0} \frac{x + \sin x}{x}$,

$= {\lim}_{x \to 0} \left\{\frac{x}{x} + \sin \frac{x}{x}\right\}$,

$= {\lim}_{x \to 0} \left\{1 + \sin \frac{x}{x}\right\}$,

$= 1 + 1$

$= 2$.

May 22, 2018

$\textcolor{b l u e}{{\lim}_{x \rightarrow 0} \frac{x}{x} + {\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1 + 1 = 2}$

#### Explanation:

Note that:

color(red)[lim_(nrarr0)sinn/n=1

${\lim}_{x \rightarrow 0} \frac{x + \sin \left(x\right)}{x}$

${\lim}_{x \rightarrow 0} \frac{x}{x} + {\lim}_{x \rightarrow 0} \sin \frac{x}{x}$

${\lim}_{x \rightarrow 0} \frac{x}{x} = \frac{0}{0}$

since the direct compensation product equal $\frac{0}{0}$
we will use L'hospital Rule.

L'hospital Rule $\textcolor{red}{{\lim}_{t \rightarrow a} \frac{f ' \left(x\right)}{g \left(' x\right)}}$

${\lim}_{x \rightarrow 0} \frac{x}{x} = {\lim}_{x \rightarrow 0} \frac{1}{1} = 1$

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

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

$\textcolor{b l u e}{{\lim}_{x \rightarrow 0} \frac{x}{x} + {\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1 + 1 = 2}$