Solve limit with square root signs I have an homework question that I can't solve. Solve the following limit: 
$$
\lim_{x \to \infty}\;(x^2+3)^\frac {1}{2}\ - (x^2 + x)^\frac {1}{2}\  
$$
 A: 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
$$
\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{3-x}{\sqrt{x^2+3}+\sqrt{x^2+x}}=\lim_{x\to\infty}\frac{\frac{3}{x}-1}{\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x}}}=-\frac12.
$$
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\root{x^{2} + 3} - \root{x^{2} + x}
=
{\pars{\root{x^{2} + 3} - \root{x^{2} + x}}\pars{\root{x^{2} + 3} + \root{x^{2} + x}} \over \root{x^{2} + 3} + \root{x^{2} + x}}
\\[3mm]&=
{\pars{x^{2} + 3} - \pars{x^{2} + x} \over \root{x^{2} + 3} + \root{x^{2} + x}}
=
-\,{x - 3 \over \root{x^{2} + 3} + \root{x^{2} + x}}
=
-\,{1 - 3/x \over \root{1 + 3/x^{2}} + \root{1 + 1/x}}
\end{align}

$$
\lim_{x \to \infty}\pars{\root{x^{2} + 3} - \root{x^{2} + x}}
=
\lim_{x \to \infty}\pars{-\,{1 - 3/x \over \root{1 + 3/x^{2}} + \root{1 + 1/x}}}
=-\,{1 \over 1 + 1}
$$

$$\color{#0000ff}{\large%
\lim_{x \to \infty}\pars{\root{x^{2} + 3} - \root{x^{2} + x}}
=
-\,\half}$$
