How do I find $\lim\limits_{x \to 0} x\cot(6x) $ without using L'Hôpital's rule? How do I find $\lim\limits_{x \to 0} x\cot(6x) $ without using L'Hôpital's 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'Hôpital's rule.
Here is how I solved it (and got the wrong answer). Hopefully someone could tell me where I went wrong.
$$\lim\limits_{x \to 0} x\cot(6x) = (\lim\limits_{x \to 0} x) (\lim\limits_{x \to 0}\cot(6x)) = (0)\left(\lim\limits_{x \to 0}\frac{\cos(6x)}{\sin(6x)}\right) = (0)(\frac{1}{0}).$$ 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'Hôpital's rule.

$$
x\cot (6x) = \frac 1 6 \cdot \frac{6x}{\sin(6x)}\cdot\cos(6x) = \frac 1 6 \cdot \frac{u}{\sin u}\cdot\cos(6x)
$$
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).
