Squeeze Theorem Problem

I assume you meant $$\lim_{x \to \infty} \dfrac{2x^3 + \sin(x^2)}{1+x^3}$$ Note that $-1 \leq \sin(\theta) \leq 1$. Hence, we have that $$\dfrac{2x^3 - 1}{1+x^3} \leq \dfrac{2x^3 + \sin(x^2)}{1+x^3} \leq \dfrac{2x^3 + 1}{1+x^3}$$ Note that $$\dfrac{2x^3 - 1}{1+x^3} = \dfrac{2x^3 +2 -3}{1+x^3} = 2 - \dfrac3{1+x^3}$$ $$\dfrac{2x^3 + 1}{1+x^3} = \dfrac{2x^3 + 2 - 1}{1+x^3} = 2 - \dfrac1{1+x^3}$$ Hence, $$2 - \dfrac3{1+x^3} \leq \dfrac{2x^3 + \sin(x^2)}{1+x^3} \leq 2 - \dfrac1{1+x^3}$$ Can you now find the limit?

EDIT

If you want to make use of the fact that $\lim_{x \to \infty} \dfrac{\sin(x^2)}{x^3} = 0$, divide the numerator and denominator of $\dfrac{2x^3 + \sin(x^2)}{1+x^3}$ by $x^3$ to get $$\dfrac{2x^3 + \sin(x^2)}{1+x^3} = \dfrac{2 + \dfrac{\sin(x^2)}{x^3}}{1 + \dfrac1{x^3}}$$ Now make use of the fact that $\lim_{x \to \infty} \dfrac{\sin(x^2)}{x^3} = 0$ and $\lim_{x \to \infty} \dfrac1{x^3} = 0$ to get your answer.

Break the function into smaller pieces:

$$\frac{2x^3+\sin x^2}{1+x^3}=\frac{2x^3}{1+x^3}+\frac{\sin x^2}{1+x^3}\;.$$

I expect that you already have tools to deal with $\lim\limits_{x\to\infty}\frac{2x^3}{1+x^3}$, and $\lim\limits_{x\to\infty}\frac{\sin x^2}{1+x^3}$ can be evaluated easily on the basis of the first part of the problem.