Answer
$1$
Work Step by Step
$\dfrac {x-1}{x}\leq \dfrac {x-\cos x}{x}\leq \dfrac {x+1}{x}$
$\lim _{x\rightarrow \infty }\dfrac {x-1}{x}=1-\dfrac {1}{x}=1-0=1$
$\lim _{x\rightarrow \infty }\dfrac {x+1}{x}=1+\dfrac {1}{x}=1+0=1$
$\Rightarrow 1\leq \lim _{x\rightarrow \infty }\dfrac {x+\cos x}{x}\leq 1\Rightarrow \lim _{x\rightarrow \infty }\dfrac {x+\cos x}{x}=1$
