For all $n\in \mathbb{N}$, let ${f}_{n}:(0,\mathrm{\infty})\to \mathbb{R}$ be the function defined by ${f}_{n}(x)=\frac{n\mathrm{sin}(x/n)}{x({x}^{2}+1)}$. Find the pointwise limit of $({f}_{n})$

logiski9s
2022-07-07
Answered

For all $n\in \mathbb{N}$, let ${f}_{n}:(0,\mathrm{\infty})\to \mathbb{R}$ be the function defined by ${f}_{n}(x)=\frac{n\mathrm{sin}(x/n)}{x({x}^{2}+1)}$. Find the pointwise limit of $({f}_{n})$

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Jamarcus Shields

Answered 2022-07-08
Author has **17** answers

The pointwise limit f of the sequence $\{{f}_{n}{\}}_{n\in {\mathbb{Z}}^{+}}$ you get by computing, for every $x>0$, the limit

$f(x)=\underset{n\to \mathrm{\infty}}{lim}{f}_{n}(x)=\underset{n\to \mathrm{\infty}}{lim}\frac{n\mathrm{sin}(x/n)}{x({x}^{2}+1)}.$

Notice that (substituting $h=\frac{1}{n}$) this is the same limit as$f(x)=\underset{h\to 0}{lim}\frac{\mathrm{sin}(xh)}{xh}\cdot \frac{1}{{x}^{2}+1}=\frac{1}{{x}^{2}+1},$

and, so as you can see, we get the same limit that you computed.

$f(x)=\underset{n\to \mathrm{\infty}}{lim}{f}_{n}(x)=\underset{n\to \mathrm{\infty}}{lim}\frac{n\mathrm{sin}(x/n)}{x({x}^{2}+1)}.$

Notice that (substituting $h=\frac{1}{n}$) this is the same limit as$f(x)=\underset{h\to 0}{lim}\frac{\mathrm{sin}(xh)}{xh}\cdot \frac{1}{{x}^{2}+1}=\frac{1}{{x}^{2}+1},$

and, so as you can see, we get the same limit that you computed.

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