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.
 A: It is true that both $a/n$ and $b/n$ tend to $0$ and $n\to\infty$, however the factor of $n$ in that term is approaching $\infty$ at the same time. So, analyzing this way, the second term gives the form $\infty\cdot 1$ and so the entire expression gives the indeterminate form $\infty-\infty$.
A: 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 $2$.
A: Your initial approach was fine.  Just use the expansion $\sqrt{1+\frac{a}{n}}=1+\frac{a}{2n}+O\left(\frac{1}{n^2}\right)$ to show that 
$$\begin{align}
\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}&=\left(1+\frac{a}{2n}+O\left(\frac{1}{n^2}\right)\right)\left(1+\frac{b}{2n}+O\left(\frac{1}{n^2}\right)\right)\\\\\
&=1+\frac{a+b}{2n}+O\left(\frac{1}{n^2}\right)
\end{align}$$
Finally we have 
$$\begin{align}
\lim_{n\to \infty}\left(n-\sqrt{n+a}\sqrt{n+b}\right)&=\lim_{n\to \infty}\,n\,\left( 1-1-\frac{a+b}{2n}+O\left(\frac{1}{n^2}\right)\right)\\\\
&=-\frac{a+b}{2}\end{align}$$
A: You can only write $$\lim_{n\to\infty}a_n-b_n=\lim_{n\to\infty}a_n-\lim_{n\to\infty}b_n$$ when both $\lim_{n\to\infty}a_n$ and $\lim_{n\to\infty}b_n$ exist. Note, that $\lim_{n\to\infty}a_n=\infty$ means, that $(a_n)$ does not converges.
A: 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*}
A: 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 $$\lim_{n \to \infty}\left(n - \sqrt{n+a}\sqrt{n+b}\right) =-\frac12(a+b). $$ 
A: Another approach to solving this is via L'Hopital's rule. For this, we first need to do a bit of rearranging...
$$
\lim_{n\to\infty} \frac{1-\sqrt{1+a/n}\sqrt{1+b/n}}{1/n}
$$
which, letting $m=1/n$, gives us
$$
\lim_{m\to0} \frac{1-\sqrt{1+am}\sqrt{1+bm}}{m}
$$
It is from here that L'Hopital's rule (in this case, it's equivalent to evaluating the derivative of $1-\sqrt{1+ax}\sqrt{1+bx}$ at $x=0$) can be applied, namely
$$
\lim_{m\to0} \frac{-\frac12\left(\frac{a\sqrt{1+bm}}{\sqrt{1+am}}+\frac{b\sqrt{1+am}}{\sqrt{1+bm}}\right)}{1} = -\frac{a+b}2
$$
