<s>[INST] 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.
 [/INST] one way to see this is to use the binomial theorem. here is how it goes:
$$(n + a)^{1/2} = n^{1/2} + \frac12n^{-1/2}a + \cdots\\
\sqrt{n+a}\sqrt{n+b} = \left(\sqrt n + \frac a{2\sqrt n}+\cdots\right) \left(\sqrt n + \frac b{2\sqrt n}+\cdots\right) = n + \frac12(a+b)+\cdots   $$
therefore $$\lim_{n \to \infty}\left(n - \sqrt{n+a}\sqrt{n+b}\right) =-\frac12(a+b). $$ 
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