Solution:
$\lim\limits_{x\to 0}\dfrac{\sqrt{x^3 + 1} - 1}{x^2 + x} = 0$
Step by step solution:
$\quad \lim\limits_{x\to 0}\dfrac{\sqrt{x^3 + 1} - 1}{x^2 + x}$
By using conjugate multiplication, we get:
$\quad \lim\limits_{x\to 0}\dfrac{\left(\sqrt{x^3 + 1} - 1\right)\left(\sqrt{x^3 + 1} + 1\right)}{(x^2 + x)\left(\sqrt{x^3 + 1} + 1\right)}$
$= \lim\limits_{x\to 0}\dfrac{x^3}{(x^2 + x)\left(\sqrt{x^3 + 1} + 1\right)}$
$= \lim\limits_{x\to 0}\dfrac{x^2}{(x + 1)\left(\sqrt{x^3 + 1} + 1\right)}$
$=\dfrac{0^2}{(0+1)\left(\sqrt{0^3 + 1} + 1\right)}$
$= 0$
$\boxed{\begin{array}{l}\\\quad Answer: \ 0\quad\\\end{array}\\}$
