What is the correct solution for the limit $\underset{x\to \mathrm{\infty}}{lim}x\mathrm{sin}\frac{1}{x}$?

Kiana Dodson
2022-06-30
Answered

What is the correct solution for the limit $\underset{x\to \mathrm{\infty}}{lim}x\mathrm{sin}\frac{1}{x}$?

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Turynka2f

Answered 2022-07-01
Author has **17** answers

$\underset{x\to \mathrm{\infty}}{lim}x\mathrm{sin}\frac{1}{x}=\underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{sin}\left(\frac{1}{x}\right)}{\frac{1}{x}}=\underset{u\to 0}{lim}\frac{\mathrm{sin}(u)}{u}=1$

Garrett Black

Answered 2022-07-02
Author has **5** answers

The first line of reasoning is clearly incorrect: let f(x)=x, $g(x)=({x}^{2}+1{)}^{-1}$. Then

$\underset{x\to \mathrm{\infty}}{lim}f(x)=\mathrm{\infty}$

and $g(x)\in (0,1]$ is clearly bounded on $\mathbb{R}$. But

$\underset{x\to \mathrm{\infty}}{lim}f(x)g(x)=\underset{x\to \mathrm{\infty}}{lim}\frac{1}{x+1/x}=0.$

$\underset{x\to \mathrm{\infty}}{lim}f(x)=\mathrm{\infty}$

and $g(x)\in (0,1]$ is clearly bounded on $\mathbb{R}$. But

$\underset{x\to \mathrm{\infty}}{lim}f(x)g(x)=\underset{x\to \mathrm{\infty}}{lim}\frac{1}{x+1/x}=0.$

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