$\infty - \infty = 0$ ? I am given this sequence with square root. $a_n:=\sqrt{n+1000}-\sqrt{n}$. I have read that sequence converges to $0$, if $n \rightarrow \infty$. Then I said, well, it may be because $\sqrt{n}$ goes to $\infty$, and then $\infty - \infty = 0$. Am I right? If I am right, why am I right? I mean, how can something like $\infty - \infty = 0$ happen, since the first $\infty$ which comes from $\sqrt{n+1000}$ is definitely bigger than $\sqrt{n}.$?
 A: But this is not really $\infty-\infty$. This is the limit of differences between two unbounded sequences. This is the key point here.
The difference is approaching zero, because $1{,}000$ is very small compared to $\sqrt n$. At some point, $\sqrt n$ is almost the size of $\sqrt{n+1000}$. For example for $n=10^{1000}$ the difference gets very small, that is $a_{10^{1000}}$ is a very small number.
A: You have to look at the sequence in particular.  In your case, a little algebra clears things up:
$$ \sqrt{n+1000} - \sqrt{n} = \frac{1000}{\sqrt{n+1000} + \sqrt{n}} \approx \frac{1000}{2 \sqrt{n}} $$
which obviously goes to zero as $n \rightarrow \infty$.
A: $$\sqrt{n+100}-\sqrt n=\frac{100}{\sqrt{n+100}+\sqrt n}\xrightarrow [n\to\infty]{}0$$
But you're not right, since for example
$$\sqrt n-\sqrt\frac{n}{2}=\frac{\frac{n}{2}}{\sqrt n+\sqrt\frac{n}{2}}\xrightarrow [n\to\infty]{}\infty$$
In fact, "a difference $\,\infty-\infty\,$ in limits theory can be anything
A: Note that 
$$\sqrt{n}\lt \sqrt{n+1000}\lt \sqrt{n}+\frac{500}{\sqrt{n}}.$$
This is because 
$$\left(\sqrt{n}+\frac{500}{\sqrt{n}}\right)^2=n+1000+\frac{500^2}{n}\gt n+1000.$$
If follows that
$$0\lt \sqrt{n+1000}-\sqrt{n}\lt \frac{500}{\sqrt{n}}.$$
Now by Squeezing we can conclude that
$$\lim_{n\to\infty}(\sqrt{n+1000}-\sqrt{n})=0.$$
A: The argument $\infty-\infty=0$ will fail you if you consider $a_n=(n+1)-n$ where the limit is obviously $1$ or $n^2-n$ where the limit is $\infty$. Indeed if $a_n\to \infty$ and $b_n\to \infty$ you can say nothing for the covergence of $a_n-b_n$ (unless you have additional clues).
In our case, although $\sqrt{n+a}$ is strictly bigger than $\sqrt{n}$, both diverge to $+\infty$ in exactly the same way. In fact, $$\sqrt{n+a}-\sqrt{n}=\frac{n+a-n}{\sqrt{n+a}+\sqrt{n}}\to a\frac{1}{\infty}=0$$
for any $a\in \mathbb{R}$. For the first equality I used the well known identity:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$$
A: No, $\infty-\infty$ is indeterminate.
This is what you can do:
$$
\sqrt{n+1000}-\sqrt{n}=\frac{(\sqrt{n+1000}-\sqrt{n})(\sqrt{n+1000}+\sqrt{n})}{\sqrt{n+1000}+\sqrt{n}}=\frac{n+1000-n}{\sqrt{n+1000}+\sqrt{n}}.
$$
Now you surely can see why this converges to $0$.
A: I think you need to revise your definition of limit of a sequence.
$\sqrt{n+1000}$ comes arbitrarily close to $\sqrt{n}$. More precisely, for any arbitrarily small $\epsilon > 0$ there exists an integer $N_\epsilon$ such that,
\begin{equation}
\sqrt{n+1000}-\sqrt{n} < \epsilon \;\; \forall \; n>N_\epsilon
\end{equation}
