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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.

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  • $\begingroup$ $\frac{a}{\infty}=0$ $\endgroup$ – hamam_Abdallah Oct 19 '16 at 18:00
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Hint :
Use Squeeze Principle. $0 \le \dfrac{1}{\sqrt{n^2 +3} + \sqrt{n^2 +2}} \le \dfrac{1}{2n}$.

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  • $\begingroup$ Thanks. I knew I could use it, its just I couldn't really come up with anything. In this case, how did you come up with $1/2n$? Do you simply know this from before or is there some inequality that I am not aware of? $\endgroup$ – Ruslan Mushkaev Oct 19 '16 at 19:28
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    $\begingroup$ @RuslanMushkaev Note that $n^2+2>n^2$ and $n^2+3>n^2$. So, $\sqrt{n^2+2}+\sqrt{n^2+3}>2n$. Finally, $$\frac{1}{\sqrt{n^2+2}+\sqrt{n^2+3}}<\frac1{2n}$$which is the same inequality that I used in the $\delta-\epsilon$ proof I posted herein. -Mark $\endgroup$ – Mark Viola Oct 19 '16 at 20:12
  • $\begingroup$ You're welcome. @Ruslan Mushkaev. To use squeeze principle, it is a standard trick - Put away all constants aside. And make the expression constant-free. You can do this by decreasing denominator and increasing numerator. $\endgroup$ – Hardey Pandya Oct 20 '16 at 2:03
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Note that for any given $\epsilon>0$, we have

$$\begin{align} \left|\sqrt{n^2+3}-\sqrt{n^2+2}\right|&=\left|\frac{1}{\sqrt{n^2+3}+\sqrt{n^2+2}}\right|\\\\ &\le \frac{1}{2n}\\\\ &<\epsilon \end{align}$$

whenever $n>N=1+\lfloor \frac2\epsilon\rfloor$.

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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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