$\infty - \infty = 0$ ?

Solution 1:

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: $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$$

Solution 2:

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 $0$.