Answer
$\dfrac{5}{3}$
Work Step by Step
We know that:
(a) $\lim\limits_{x \to a} k(x)=k(a)$ where $a$ is a constant.
(b) $\lim\limits_{x \to a} \dfrac{p(x)}{q(x)}=\dfrac{\lim\limits_{x \to a} p(x)}{\lim\limits_{x \to a} q(x)}$
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Thus, we have:
$\lim\limits_{x \to 2} \dfrac{3x+4}{x^2+x}=\dfrac{\lim\limits_{x \to 2} 3x+4}{\lim\limits_{x \to 2} x^2+x}=\dfrac{(3)(2)+4}{(2)^2+4}=\dfrac{5}{3}$
