<s>[INST] $\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$ - 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?
 [/INST] $$\sqrt{2^n+n^2}-\sqrt{2^n+1}=2^{n/2} (\sqrt{1+n^22^{-n}}-\sqrt{1+2^{-n}})=2^{n/2}\left(1+\frac12n^22^{-n}+o(n^22^{-n})-1-\frac122^{-n}-o(2^{-n})\right)=\frac{n^2-1}22^{-n/2}+o(n^22^{-n/2}).$$
The expression converges to $0$.
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