<s>[INST] Compute $ \sqrt{n^2 +3} - \sqrt{n^2 +2} $. I have a question concerning determining a basic limit of $ \sqrt{n^2 +3} - \sqrt{n^2 +2} $ as $n \rightarrow \infty $. Upon using the difference of squares, this yields: $ \frac{1}{\sqrt{n^2 +3} + \sqrt{n^2 +2}} $Now the limit seems intuitive, but I cant find a formal way of determining it. Any hints about what the next step should be would be appreciated.
 [/INST] Let $f(x) = \sqrt{1+3x} -\sqrt{1+2x}$.
Then $f(0) = 0$ and $f'(0) = {1 \over 2}$, hence for $x$ sufficiently
small, we have $|f(x)| \le |x|$.
Hence for $n$ sufficiently large, we have
$|\sqrt{n^2+3} -\sqrt{n^2+2} | = n |f({1 \over n^2})| \le {1 \over n}$,
from which the limit follows.
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