# Evaluate the following limits. lim_{nrightarrowinfty}n^2ln(nsinfrac{1}{n})

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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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 LHospital