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
$$\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$$
 A: I will assume you know that $\frac{\sin u}{u}\xrightarrow[u\to 0]{} \sin^\prime(0) = 1$.


*

*First, observe that $\frac{x+1}{4x^2+3} \xrightarrow[x\to\infty]{} 0$.

*Use this observation to rewrite
$$
\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)
= \frac{2x^3+x}{x^2+1}\frac{x+1}{4x^2+3} \frac{\sin\left(\frac{x+1}{4x^2+3}\right)}{\frac{x+1}{4x^2+3}}
$$

*Compute the limit of the first term, now that the second is dealt with:
$$
\frac{2x^3+x}{x^2+1}\frac{x+1}{4x^2+3} = \frac{2x}{4x}\frac{2+x^{-2}}{1+x^{-2}}\frac{1+x^{-1}}{4+3x^{-2}} \xrightarrow[x\to\infty]{}\frac{1}{2}
$$

*Conclude that
$$
\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right) \xrightarrow[x\to\infty]{}\frac{1}{2} \cdot 1 = \frac{1}{2}.
$$
A: 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 $x\to \infty$.
