$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$ - is there any other way to solve this? Here's a problem that I think I've managed to solve:
$$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$$
Here's how I did it:
$(\sqrt{2^n+n^2}-\sqrt{2^n+1})\frac{\sqrt{2^n+n^2}+\sqrt{2^n+1}}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}=\frac{n^2-1}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}$
And now you can see that (for sufficiently large $n$):
$0\le\frac{n^2-1}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}\le\frac{n^2}{\sqrt{2^n}}$
Limit of $0$ clearly is $0$, and this is also true for $\frac{n^2}{\sqrt{2^n}}$, so that means that:
$$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}=0$$
The thing is I got kind of lucky I've solved it and I'm interested if there are any other ways to solve this limit (not involving any derivatives nor integrals). Does anybody see any clever solution other than mine? And also - is my solution correct?

If you know about Taylor expansions, you can do the following: as long as $\frac{a_n}{2^n}\xrightarrow[n\to\infty]{}0$,
$$
\sqrt{2^n + a_n} = \sqrt{2^n}\sqrt{1 + \frac{a_n}{2^n}} = \sqrt{2^n}\left(1+\frac{a_n}{2\cdot2^n}+o\!\left(\frac{a_n}{2^n}\right)\right) = 2^{\frac{n}{2}}+\frac{a_n}{2^{\frac{n}{2}+1}}+o\!\left(\frac{a_n}{2^n}\right)
$$
Applying this to your case with first $a_n=n^2$, then $a_n=1$, you get that the difference is
$$
2^{\frac{n}{2}}+\frac{n^2}{2^{\frac{n}{2}+1}}+o\!\left(\frac{n^2}{2^n}\right) - 2^{\frac{n}{2}}
$$
as $\frac{1}{2^n} = o\!\left(\frac{n^2}{2^n}\right)$. I.e., the difference is equivalent to $\frac{n^2}{2^{\frac{n}{2}+1}}$, which goes to 0.
