Answer
$\infty$
Work Step by Step
consider $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} (\sqrt {x^3+1}-\sqrt x)$
Re-arrange as:
$\lim\limits_{x \to \infty} (\sqrt {x^3+1}-\sqrt x)=\lim\limits_{x \to \infty} x (\dfrac{\sqrt {x^3+1}}{x}-\dfrac{\sqrt x}{x})$
Thus, $\lim\limits_{x \to \infty} (\sqrt {1+\dfrac{1}{x^2}}-\sqrt{\dfrac{1}{x}})=\lim\limits_{x \to \infty} (x) \cdot (1)$
and $\lim\limits_{x \to \infty} (x)(1)=\infty$
