Answer
$\lim\limits_{x\to2}\dfrac{x}{x^2+1}=\dfrac{2}{5}.$
Work Step by Step
By Theorem $1.3: \lim\limits_{x\to c}r(x)=r(c)=\dfrac{p(c)}{q(c)}$ where $r(x)=\dfrac{p(x)}{q(x)}.$
$\lim\limits_{x\to2}\dfrac{x}{x^2+1}=\dfrac{2}{2^2+1}=\dfrac{2}{5}.$