How to evaluate $\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$$

Hint:
\begin{align}
\lim_{x\to\infty}\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)&=\left(\lim_{x\to\infty}\frac{(2x^3+x)(x+1)}{(x^2+1)(4x^2+3)}\right)\lim_{x\to\infty}\frac{\sin\left(\frac{x+1}{4x^2+3}\right)}{\frac{x+1}{4x^2+3}}\\
\end{align}
Now, by making $\color{blue}{t=\frac{x+1}{4x^2+3}}$, $t\to 0^+$ as $x\to \infty$.
