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How do I find $\lim\limits_{x \to 0} x\cot(6x) $ without using L'hopitals rule?

I'm a freshman in college in the 3rd week of a calculus 1 class. I know the $\lim\limits_{x \to 0} x\cot(6x) = \frac{1}{6}$ by looking at the graph, but I'm not sure how to get here without using L'hopital's rule.

Here is how I solved it (and got the wrong answer). Hopefully someone could tell me where I went wrong.

$\lim\limits_{x \to 0} x\cot(6x) = (\lim\limits_{x \to 0} x) (\lim\limits_{x \to 0}cot(6x)) = (0)((\lim\limits_{x \to 0}\frac{cos(6x)}{sin(6x)}) = (0)(\frac{1}{0}). $ Therefore the limit does not exist.

I am also unsure of how to solve $\lim\limits_{x \to 0} \frac{\sin(5x)}{7x^2} $ without using L'hopital's rule.

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  • $\begingroup$ Are you allowed to use the limit for $\frac{\sin x}x$? $\endgroup$ – Mark Bennet Sep 14 '14 at 17:55
  • $\begingroup$ yes that is all I am allowed to use. $\endgroup$ – Alex Sep 14 '14 at 17:56
  • $\begingroup$ Well others have shown in their answers how that will help, but you still need to be a little bit careful to work accurately with the limits you know and not just to assume that you can take one limit inside another one (this is often possible, and there are theorems about it, but if you are working from first principles you should make sure it works in these cases). $\endgroup$ – Mark Bennet Sep 14 '14 at 18:06
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HINT:

$$x\cot6x=\frac16\frac{6x}{\sin6x}\cdot\cos6x$$

$$\frac{\sin5x}{x^2}=5\frac{\sin5x}{5x}\cdot\frac1x$$

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$$ x\cot (6x) = \frac 1 6 \cdot \frac{6x}{\sin(6x)}\cdot\cos(6x) = \frac 1 6 \cdot \frac{u}{\sin u}\cdot\cos(6x) $$ and as $x$ approaches $0$, so does $u$. The fraction $u/\sin u$ has both the numerator and denominator approaching $0$, and its limit is well known to be $1$ (presumably you've seen that one before).

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