<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] Your error is in this equality:
\begin{equation*}
\lim\limits_{n\rightarrow \infty }n-n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}%
=0\ \ \ \ \ \ (Here\ it\ is).
\end{equation*}
Explanation. You known that 
\begin{eqnarray*}
\lim_{n\rightarrow \infty }n &=&+\infty ,\ \ \ and \\
\lim_{n\rightarrow \infty }n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}
&=&+\infty ,\ \ \ too.
\end{eqnarray*}
However, it seems that you forget that $\infty -\infty $ is an indetermined
form, that is, when one have
\begin{equation*}
\lim_{n\rightarrow \infty }a_{n}=\infty ,\ \ \ \ \ and\ \ \ \ \ \
\lim_{n\rightarrow \infty }b_{n}=\infty 
\end{equation*}
then one cannot give any conclusion about
\begin{equation*}
\lim_{n\rightarrow \infty }(a_{n}-b_{n})
\end{equation*}
because one can have many situations. In this case one should make some
further study. For this example you use the classic technique which consists
of multiplying and dividing by what is called the conjugate of the original
expression, as follows (step by step)
\begin{eqnarray*}
\left( n-n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}\right)  &=&\left( n-n%
\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}\right) \left( \frac{n+n\sqrt{1+%
\frac{a}{n}}\sqrt{1+\frac{b}{n}}}{n+n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}%
}\right)  \\
&=&\frac{n^{2}-\left( n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}\right) ^{2}}{%
n+n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}},\ \ \ \ \ \
((x-y)(x+y)=x^{2}-y^{2}) \\
&=&\frac{n^{2}-n^{2}(1+\frac{a}{n})(1+\frac{b}{n})}{n+n\sqrt{1+\frac{a}{n}}%
\sqrt{1+\frac{b}{n}}} \\
&=&\frac{n^{2}(1-(1+\frac{a}{n})(1+\frac{b}{n}))}{n(1+\sqrt{1+\frac{a}{n}}%
\sqrt{1+\frac{b}{n}})} \\
&=&\frac{n(1-\left( 1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^{2}}\right) )}{(1+%
\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}})} \\
&=&\frac{-(a+b+\frac{ab}{n})}{(1+\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}})}
\end{eqnarray*}
therefore, passing to the limit one gets
\begin{eqnarray*}
\lim_{n\rightarrow \infty }\left( n-n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}%
\right)  &=&\lim_{n\rightarrow \infty }\frac{-(a+b+\frac{ab}{n})}{(1+\sqrt{1+%
\frac{a}{n}}\sqrt{1+\frac{b}{n}})} \\
&=&\frac{-(a+b+0)}{1+\sqrt{1+0}\sqrt{1+0}}=-\frac{(a+b)}{2}
\end{eqnarray*}
</s>