Squeeze Theorem Problem I'm busy studying for my Calculus A exam tomorrow and I've come across quite a tough question. I know I shouldn't post such localized questions, so if you don't want to answer, you can just push me in the right direction.
I had to use the squeeze theorem to determine:
$$\lim_{x\to\infty} \dfrac{\sin(x^2)}{x^3}$$
This was easy enough and I got the limit to equal 0. Now the second part of that question was to use that to determine:
$$\lim_{x\to\infty} \dfrac{2x^3 + \sin(x^2)}{1 + x^3}$$
Obvously I can see that I'm going to have to sub in the answer I got from the first limit into this equation, but I can't seem to figure how how to do it.
Any help would really be appreciated! Thanks in advance!

First of all we know that, $-1 \le \sin x^2 \le 1$.
Step 1: We should add $2x^3$ on all sides of inequalities so that we obtain $$2x^3-1 \le 2x^3 + \sin x^2 \le 2x^3+1$$ as our new inequality.
Step 2: As $x$ tends to $+\infty$, it is reasonable to assume that $1+x^3>0$, so we divide all sides of the inequalities by $1+x^3$ without reversing the inequalities, so we obtain $$\frac {2x^3-1} {1+x^3} \le \frac {2x^3 + \sin x^2} {1+x^3} \le \frac {2x^3+1} {1+x^3}$$ as our new inequality.
Step 3: Apply limit on both extreme left hand side and extreme right side with $x \to \infty$. You will definitely get the answer $2$.
So $\lim _{x \to \infty} \dfrac {2x^3 + \sin x^2} {1+x^3} = 2$. That's it.
