How to evaluate $\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$?

Question

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 $$\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$$

Answer 1

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

Answer 2

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 $x\to \infty$.