What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}}$ and so: $$\lim \limits _ {n \to \infty} n - n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}} = 0$$
Because both $\frac{a}{n}$ and $\frac{b}{n}$ tend to $0$.
What would give a correct answer:
Plotting the function $$f(x) = x - \sqrt{x+a} \sqrt{x+b}$$
Clearly indicates that it has an asymptote in $- \frac{a+b}{2}$. This result can be obtained multiplying the numerator and the denominator by $n + \sqrt{n+a} \sqrt{n+b}$:
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = $$
$$-\lim \limits _{n \to \infty} \frac {n(a+b)}{n + \sqrt{n+a} \sqrt{n+b}} - \lim \limits _{n \to \infty} \frac {ab}{n + \sqrt{n+a} \sqrt{n+b}}$$
The second limit is clearly $0$ and the first one gives the correct answer (dividing the numerator and denominator by $n$).
Why the first way I tried is wrong? I might have done something silly but I cannot find it.

Another approach:
$$\sqrt{a+n}\sqrt{b+n}=\sqrt{ab+(a+b)n+n^2}$$ and 
$$\begin{align}n-\sqrt{n+a}\sqrt{n+b}&=\frac{n^2-(n+a)(n+b)}{n+\sqrt{n^2+(a+b)n+ab}}\\
&=\frac{-(a+b)n-ab}{n+\sqrt{n^2+(a+b)n+ab}}\\
&=\frac{-(a+b)-\frac{ab}{n}}{1 + \sqrt{1+\frac{a+b}{n} + \frac{ab}{n^2}}}
\end{align}$$
The numerator converges to $-(a+b)$ and the denominator converges to $2$.
