limit of the sequence $a_n=n\left[\sin\left(\frac{1+n^3}{n^2}\right)-\sin n\right]$ I started by writting
$$
\begin{eqnarray*}
a_n &=& n \left[\sin\left(\frac{1+n^3}{n^2}\right)-\sin\, n\right]=\frac{\sin\left(\displaystyle\frac{1}{n^2}+n\right)-\sin\, n}{\displaystyle\frac{1}{n}} \\
    &=&\frac{2\sin\left(\displaystyle\frac{1}{n^2}\right)\cos\left(\displaystyle\frac{1}{n^2}+2n\right)}{\displaystyle\frac{1}{n}}
\end{eqnarray*}
$$
after putting 2 in the denominator of the denominator, I got stuck.
Solution's manual of the book follows my reasoning but writting the result after using the trigonometric identity was different and I did not understood it. Did I do something wrong?


The basic formula is $sin(x+y)-sin(x-y)=2cos(x)sin(y)$.   Here $x+y=n+\frac{1}{n^2}$ and $x-y=n$, resulting in $x=n+\frac{1}{2n^2}$ and $y=\frac{1}{2n^2}$. So $a_n=2ncos(n+\frac{1}{2n^2})sin(\frac{1}{2n^2})$ which is the expression in the book.
Your error was in getting expressions for $x$ and $y$.
