Solve limit with square root signs I have an homework question that I can't solve. Solve the following limit: 
$$
\lim_{x \to \infty}\;(x^2+3)^\frac {1}{2}\ - (x^2 + x)^\frac {1}{2}\  
$$

For every $x\in D=(-\infty,-1]\cup[0,\infty)$ we have
$$
f(x):=\sqrt{x^2+3}-\sqrt{x^2+x}=\frac{(x^2+3)-(x^2+x)}{\sqrt{x^2+3}+\sqrt{x^2+x}}=\frac{3-x}{\sqrt{x^2+3}+\sqrt{x^2+x}}.
$$
It follows that
$$
\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{3-x}{\sqrt{x^2+3}+\sqrt{x^2+x}}=\lim_{x\to\infty}\frac{\frac{3}{x}-1}{\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x}}}=-\frac12.
$$
