I want to calculate the limit of $\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x)$ I want to calculate the limit of: $$\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x) $$ $$(a,b ∈ R)$$
Now I know the result is $\frac{1}{2}(a+b)$, but I am having trouble getting to it.
 A: Write $$\sqrt{(x+a)(x+b)}-x=\frac{(a+b)x+ab}{\sqrt{(x+a)(x+b)}+x}=\frac{a+b+\frac{ab}{x}}{\sqrt{(1+\frac ax)(1+\frac bx)}+1}.$$
Now let $x\to+\infty$.
A: A standard method is
\begin{align}
\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x)
&=
\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x)
  \frac{\sqrt{(x+a)(x+b)}+x}{\sqrt{(x+a)(x+b)}+x}\\
&=
\lim_{x \to +\infty} \frac{(a+b)x+ab}{\sqrt{(x+a)(x+b)}+x}\\
&=
\lim_{x \to +\infty} \frac{(a+b)+\frac{ab}{x}}{\sqrt{(1+\frac{a}{x})(1+\frac{b}{x})}+1}\\
\end{align}
A perhaps less standard method is to use the substitution $t=1/x$ so the limit becomes
$$
\lim_{t\to0^+}\frac{\sqrt{(1+at)(1+bt)}-1}{t}
$$
and you recognize the derivative at $0$ of the function
$$
f(t)=\sqrt{(1+at)(1+bt)}=\sqrt{1+(a+b)t+abt^2}
$$
Since
$$
f'(t)=\frac{(a+b)+2abt}{2\sqrt{1+(a+b)t+abt^2}}
$$
you're done.
A: Just another way to do it using Taylor expansions.
Since $x$ is large and positive $$A=\sqrt{(x+a)(x+b)}-x=\sqrt{x+a}\times\sqrt{x+b}-x=\sqrt x\times\sqrt{1+\frac a x}\times\sqrt x\times\sqrt{1+\frac b x}-x$$ $$A=x\left(\sqrt{1+\frac a x}\times\sqrt{1+\frac b x} -1\right)$$ Now, remembering that, for small values of $y$, $\sqrt {1+y} \approx 1+\frac y 2$, replace $y$ by $\frac a x$ in the first radical, then by  $\frac b x$ in the second radical, expand and go to the limit.
A: Notice, $$\lim_{x\to \infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
 $$=\lim_{x\to +\infty}\frac{\left(\sqrt{(x+a)(x+b)}-x\right)\left(\sqrt{(x+a)(x+b)}+x\right)}{\left(\sqrt{(x+a)(x+b)}+x\right)}$$
$$=\lim_{x\to +\infty}\frac{x^2+(a+b)x+ab-x^2}{\left(\sqrt{(x+a)(x+b)}+x\right)}$$
$$=\lim_{x\to +\infty}\frac{(a+b)+\frac{ab}{x}}{\left(\sqrt{\left(1+\frac{a}{x}\right)\left(1+\frac{a}{x}\right)}+1\right)}$$
$$=\frac{a+b+0}{\left(\sqrt{\left(1+0\right)\left(1+0\right)}+1\right)}$$
$$=\frac{a+b}{1+1}=\frac{a+b}{2}$$
A: Let's call the expression $A$. First note that
$$A=x((1+\frac{a+b}x+\frac{ab}{x^2})^{1/2}-1)$$
Use expansion with the little o notation 
$$A=x(1+\frac12(\frac{a+b}x+\frac{ab}{x^2}+o(\frac1x))+o(\frac1x)-1)=\frac12(a+b)+o(1)$$
as $x$ tends to infinity. 
