# Evaluate the following limits. $$\displaystyle\lim_{{{n}\rightarrow\infty}}{n}^{{2}}{\ln{{\left({n}{\sin{{\frac{{{1}}}{{{n}}}}}}\right)}}}$$

Evaluate the following limits.
$\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)$
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delai59qk
Consider the provided expression,
$\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)$
Evaluate the following limits.
We can write as,
$\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)=\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\cdot \underset{n\to \mathrm{\infty }}{lim}\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)$
Simplifying further,
Apply the chain rule of the limit.
$\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)=\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\cdot \left(\underset{n\to \mathrm{\infty }}{lim}\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)\right)$
Apply the algebraic property,
$\mathrm{sin}\left(\frac{1}{n}\right)n=\frac{\mathrm{sin}\left(\frac{1}{a}\right)}{\frac{1}{a}}$
Now, the expression becomes.
$\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)=\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\cdot \left(\underset{n\to \mathrm{\infty }}{lim}\mathrm{ln}\left(\frac{\mathrm{sin}\left(\frac{1}{a}\right)}{\frac{1}{a}}\right)\right)$
Apply L'Hospital rule,
$\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\mathrm{ln}\left(n\mathrm{sin}\frac{1}{n}\right)=\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\cdot \left(\underset{n\to \mathrm{\infty }}{lim}{\mathrm{ln}lim}_{n\to \mathrm{\infty }}\left(\frac{-\frac{\mathrm{cos}\left(\frac{1}{a}\right)}{{a}^{2}}}{-\frac{1}{{a}^{2}}}\right)\right)$
$=\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\cdot \left(\underset{n\to \mathrm{\infty }}{lim}{\mathrm{ln}lim}_{n\to \mathrm{\infty }}\left(\mathrm{cos}\left(\frac{1}{n}\right)\right)\right)$
$=\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\cdot \left(\underset{n\to \mathrm{\infty }}{lim}\mathrm{ln}1\right)\right)$
$=\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\cdot 0$
$=0$