<s>[INST] $\forall n \in \mathbb{N^*}$, prove that $|{\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}| < \frac{1}{2n^2}$ Let $$x_n = \frac{\sqrt{n^2+2}}{2n},n \in \mathbb{N^*} \ldots$$
Prove that $$|{x_n - \frac{1}{2}}| < \frac{1}{2n^2}$$
Indication: Use the relationship: $\sqrt{1+\gamma} < 1 + \frac{\gamma}{2}$, $ \forall \gamma \in \mathbb{R^{*}_{+}}$
I'd appreciate it very much if any answers could indicate a heuristic or general mindset one should have when proving such a proposition.
 [/INST] First of all we note that $\left|{\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}\right| = {\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}$ because ${\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}>0$. In fact $\frac{\sqrt{n^2+2}}{2n}>\frac{1}{2} \Rightarrow \frac{n^2+2}{n^2}>1 \Rightarrow 1+\frac{2}{n^2}>1$.
Now we note that
$\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}=\frac{\sqrt{\frac{2}{n^2}+1}}{2} - \frac{1}{2}$, and from indication $\sqrt{1+\gamma} < 1 + \frac{\gamma}{2}$, we have:
$\frac{\sqrt{\frac{2}{n^2}+1}}{2} - \frac{1}{2}<\frac{\frac{1}{n^2}+1}{2} - \frac{1}{2}=\frac{1}{2n^2}$
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