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I have four limit problems, for homework, but I don't quite understand them. I will only ask for help on one of them because that should be all I need to understand these. According to the Calculus professor, 'I do not know L'Hopital's Rule, yet.' Therefore, I may not use it L'Hopital's Rule. We have went as far as to understand $\lim_{x \to 0}$$\frac{\sin x}{x} = 1$.

The problem is:

$$\lim_{x\to 0}\frac{x^2-x}{\sin3x} $$

Thank you, for your help.

-Rux

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  • $\begingroup$ What else do you know as far as the class is concerned? Are you familiar with the sum expansion of $\sin x$? How many terms would be required to solve this problem? $\endgroup$
    – abiessu
    Commented Sep 23, 2013 at 21:35
  • $\begingroup$ take out $x$-factor common from numerator, and use this to show that the limit is $-1/3$ $\endgroup$
    – S L
    Commented Sep 23, 2013 at 21:37

3 Answers 3

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Hint: $$\frac{x^2-x}{\sin 3x} = \frac{x(x-1)}{\sin 3x} = \frac{3x}{\sin 3x} \cdot \frac{x-1}{3}$$

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  • $\begingroup$ Okay, I understand this. Thank you, for the help. I can get it from here. $\endgroup$
    – JRuxDev
    Commented Sep 23, 2013 at 21:51
  • $\begingroup$ You're welcome. $\endgroup$
    – mrf
    Commented Sep 23, 2013 at 21:55
  • $\begingroup$ Just to continue with this answer, you factor our the $\mathcal x$ from the top in the first step. Then, you can multiply the top and bottom by 3 and split that up into the final equation you have listed. I see that $\frac{3x}{\sin 3x}$ is equivalent to $\frac{\sin x}{x}$ which is 1. So the next step would be $\mathcal x $-$\frac {1}{3}$? And, so $\mathcal x = 0$, since the limit is $\lim {x \to 0}$. $\frac{0-1}{3}$ = $\frac{-1}{3}$? $\endgroup$
    – JRuxDev
    Commented Sep 23, 2013 at 22:18
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Hint: Divide the numerator and denominator by $x$

$$\lim_{x \rightarrow0} \dfrac{\frac{x^2-x}{x}}{\frac{\sin 3x}{x}}= \lim_{x \rightarrow0} \dfrac{x-1}{\frac{\sin 3x}{x}}$$

Can you use any formula for limits in the denominator now?

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Hint:
$$\lim\limits_{x\to 0} \dfrac{\sin x}{x} =\ldots$$

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