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!

  • $\begingroup$ You’re missing required parentheses in the second limit. (At least I’m assuming that the function should be $\frac{2x^3+\sin x^2}{1+x^3}$ and not, as you’ve written, $2x^3+\frac{\sin x^2}1+x^3$.) $\endgroup$ – Brian M. Scott Oct 30 '12 at 17:52
  • $\begingroup$ Yes, sorry about that. I really need to learn how to use that markup. $\endgroup$ – nickcorin Oct 30 '12 at 17:56
  • 3
    $\begingroup$ Here’s a pretty good basic MathJax tutorial. $\endgroup$ – Brian M. Scott Oct 30 '12 at 18:05

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?


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.

  • $\begingroup$ Yes, thanks for the markup. I'm not too clued up on how to make it look so nice. I thought about doing it like that, but the question implied that we should use the above answer to solve it. But thank you! I appreciate the help a lot :) $\endgroup$ – nickcorin Oct 30 '12 at 17:54
  • $\begingroup$ Absolutely perfect! Genius :) $\endgroup$ – nickcorin Oct 30 '12 at 18:00

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.

  • $\begingroup$ Awesome! :) Thank you! $\endgroup$ – nickcorin Oct 30 '12 at 17:57
  • $\begingroup$ @nickcorin: You’re welcome. $\endgroup$ – Brian M. Scott Oct 30 '12 at 17:59

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.


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