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For all $n\in \mathbb{N}$, let ${f}_{n}:\left(0,\mathrm{\infty }\right)\to \mathbb{R}$ be the function defined by ${f}_{n}\left(x\right)=\frac{n\mathrm{sin}\left(x/n\right)}{x\left({x}^{2}+1\right)}$. Find the pointwise limit of $\left({f}_{n}\right)$
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Jamarcus Shields
The pointwise limit f of the sequence $\left\{{f}_{n}{\right\}}_{n\in {\mathbb{Z}}^{+}}$ you get by computing, for every $x>0$, the limit
$f\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}{f}_{n}\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{n\mathrm{sin}\left(x/n\right)}{x\left({x}^{2}+1\right)}.$
Notice that (substituting $h=\frac{1}{n}$) this is the same limit as$f\left(x\right)=\underset{h\to 0}{lim}\frac{\mathrm{sin}\left(xh\right)}{xh}\cdot \frac{1}{{x}^{2}+1}=\frac{1}{{x}^{2}+1},$
and, so as you can see, we get the same limit that you computed.