Answer
$0$
Work Step by Step
$\dfrac {1}{2x+1}\leq \dfrac {1}{2x+\sin x}\leq \dfrac {1}{2x-1}$
$\lim _{x\rightarrow \infty }\dfrac {1}{2x+1}=\dfrac {1}{x\left( 2+\dfrac {1}{x}\right) }=\dfrac {1}{x\times \left( 2+0\right) }=\dfrac {1}{2x}=0$
$\lim _{x\rightarrow \infty }\dfrac {1}{2x-1}=\dfrac {1}{x\left( 2-\dfrac {1}{x}\right) }=\dfrac {1}{x\times \left( 2-0\right) }=\dfrac {1}{2x}=0$
$\Rightarrow 0\leq \lim _{x\rightarrow \infty }\dfrac {1}{2x+\sin x}\leq 0\Rightarrow \lim _{x\rightarrow \infty }\dfrac {1}{2x+\sin x}=0$
