Answer
(a) -1
(b) 1
(c) does not exist
Work Step by Step
$f(x)=\dfrac{x^2+x}{\sqrt {x^3+x^2}}$
First, graph the function (image attached below)
(a) As x goes to 0 from the left hand side, y approaches -1. Therefore, $\lim\limits_{x \to 0^-}f(x)=-1$
(b) As x goes to 0 from the right hand side, y approaches 1. Therefore, $\lim\limits_{x \to 0^+}f(x)=1$
(c) $\lim\limits_{x \to 0}f(x)$ does not exist because $\lim\limits_{x \to 0^-}f(x)\ne\lim\limits_{x \to 0^+}f(x)$
