Check my solution: Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$ Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$
First I rationalized the numerator,
$$
\begin{align*}
&\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x) \cdot \frac{\sqrt{(x+a_1)(x+a_2)}+x}{\sqrt{(x+a_1)(x+a_2)}+x} \\
=&\lim_{x\to+\infty}\frac{x(a_1+a_2)+a_1a_2}{\sqrt{(x+a_1)(x+a_2)}+x} \\
=&\lim_{x\to+\infty}\frac{(a_1+a_2)+(a_1a_2)/x}{\sqrt{(\frac{1}{x}+\frac{a_1}{x^2})(\frac{1}{x}+\frac{a_2}{x^2})}+1} \\
=& a_1+a_2
\end{align*}
$$
Some calculations in my calculator tells me my answer is likely to be wrong. 
I think the real answer is $\frac{a_1+a_2}{2}$.
Can anyone find my mistake or drop me a hint?
 A: Putting $h=\frac1x$
$$\lim_{h\to0}\frac{\sqrt{(1+a_1h)(1+a_2h)}-1}h$$
$$=\lim_{h\to0}\frac{(1+a_1h)(1+a_2h)-1}{h\cdot(\sqrt{(1+a_1h)(1+a_2h)}+1)}\text{( rationalizing the numerator)}$$
$$=\lim_{h\to0}\frac{a_1+a_2+a_1a_2h}{\sqrt{(1+a_1h)(1+a_2h)}+1}\text{ as }h\ne0\text{ as }h\to0$$
$$=\frac{a_1+a_2}{\sqrt{(1)(1)}+1}$$
A: You do it right until the last line $\lim_{x\to+\infty}\frac {x(a_1+a_2)+a_1a_2}{\sqrt {(x+a_1)(x+a_2)}+x}=\lim_{x\to+\infty}\frac {x(a_1+a_2)+a_1a_2}{\sqrt {x^2+(a_1+a_2)x+a_1a_2} +x}$=$\lim_{x\to+\infty}\frac {x((a_1+a_2)+a_1a_2x^{-1})}{|x| \left( \sqrt {1+(a_1+a_2)x^{-1}+a_1a_2x^{-2}}+1 \right)} $
For $x>0$ we have $|x|=x$ and thus we have $\lim_{x\to+\infty}\frac {(a_1+a_2)+a_1a_2}{\sqrt {1+(a_1+a_2)x^{-1}+a_1a_2x^{-2}}+1}=
(a_1+a_2)/2$
A: The mistake happens when you divide numerator and denominator by $x$. Inside the square root this means division by $x^{2}$ and so you should divide one factor $(x + a_{1})$ by $x$ and another factor $(x + a_{2})$ by another $x$ so that after the division the denominator will be $$\sqrt{\left(1 + \frac{a_{1}}{x}\right)\left(1 + \frac{a_{2}}{x}\right)} + 1$$ and this tends to $2$ as $x \to \infty$. You have mistakenly divided both factors inside square root by $x^{2}$ so effectively this means that the square root expression has been divided by $x^{2}$ instead of $x$.
