Squeeze Theorem Problem

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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.