Answer
$0$
Work Step by Step
$\dfrac {-1}{x}\leq \dfrac {\cos x}{x}\leq \dfrac {1}{x}$
$\lim _{x\rightarrow \infty }\dfrac {-1}{x}=\dfrac {\dfrac {d}{dx}\left( -1\right) }{\dfrac {d}{dx}\left( x\right) }=\dfrac {0}{1}=0$
$\lim _{x\rightarrow \infty }\dfrac {1}{x}=\dfrac {\dfrac {d}{dx}\left( -1\right) }{\dfrac {d}{dx}\left( x\right) }=\dfrac {0}{1}=0$
$\Rightarrow 0\leq \lim _{x\rightarrow \infty }\dfrac {\cos x}{x}\leq 0\Rightarrow \lim _{x\rightarrow \infty }\dfrac {\cos x}{x}=0$
