I know the value of the limit is $-{\displaystyle \frac{1}{6}}$. I tried to use the fact

$\mathrm{ln}(ab)=\mathrm{ln}(a)+\mathrm{ln}(b)$

$\mathrm{ln}(ab)=\mathrm{ln}(a)+\mathrm{ln}(b)$

civilnogwu
2022-07-05
Answered

I know the value of the limit is $-{\displaystyle \frac{1}{6}}$. I tried to use the fact

$\mathrm{ln}(ab)=\mathrm{ln}(a)+\mathrm{ln}(b)$

$\mathrm{ln}(ab)=\mathrm{ln}(a)+\mathrm{ln}(b)$

You can still ask an expert for help

kawiarkahh

Answered 2022-07-06
Author has **15** answers

We use

$\underset{y\to 0}{lim}\frac{\mathrm{ln}(y+1)}{y}=1$

Letting $y=n\mathrm{sin}\frac{1}{n}-1$ so

${n}^{2}\mathrm{ln}(n\mathrm{sin}\frac{1}{n})={n}^{2}(n\mathrm{sin}\frac{1}{n}-1)\frac{\mathrm{ln}(n\mathrm{sin}\frac{1}{n})}{n\mathrm{sin}\frac{1}{n}-1}$

so the limit is replaced by

$\underset{n\to \mathrm{\infty}}{lim}{n}^{2}(n\mathrm{sin}\frac{1}{n}-1)$

and as remarked this is an application of

$\underset{x\to 0}{lim}\frac{\mathrm{sin}x-x}{{x}^{3}}$

$\underset{y\to 0}{lim}\frac{\mathrm{ln}(y+1)}{y}=1$

Letting $y=n\mathrm{sin}\frac{1}{n}-1$ so

${n}^{2}\mathrm{ln}(n\mathrm{sin}\frac{1}{n})={n}^{2}(n\mathrm{sin}\frac{1}{n}-1)\frac{\mathrm{ln}(n\mathrm{sin}\frac{1}{n})}{n\mathrm{sin}\frac{1}{n}-1}$

so the limit is replaced by

$\underset{n\to \mathrm{\infty}}{lim}{n}^{2}(n\mathrm{sin}\frac{1}{n}-1)$

and as remarked this is an application of

$\underset{x\to 0}{lim}\frac{\mathrm{sin}x-x}{{x}^{3}}$

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