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

Question

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.

Answer 1

HINT:

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

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

Answer 2

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

Answer 3

Convert $\cot 6x$ to $\tan 6x$ and multiply and divide by $6$ : \begin{align} &\lim_{x \to 0} x \cot(6x) \\ = & \lim_{x \to 0} \frac{x}{\tan(6x)} \\ = & \lim_{x \to 0} \frac{6x}{6\tan(6x)} \end{align} We know that $$\lim_{x \to 0} \frac{\tan x}{x} = 1.$$

Therefore,

$$ \lim_{x \to 0} \frac{6x}{6\tan(6x)} = \frac 16\lim_{x \to 0} \cdot \frac{6x}{\tan (6x)} = \frac{1}{6}$$

Answer 4

By Taylor's formula, $$\cot(x)=\frac{\cos(x)}{\sin(x)}=\left(1-\dfrac{x^2}{2}+O(x^4)\right)\frac{1}{x-\dfrac{x^3}{6}+O(x^5)}=\left(1-\dfrac{x^2}{2}+O(x^4)\right)\frac{1}{x}\frac{1}{\left(1-\dfrac{x^2}{6}+O(x^4)\right)}$$ as $x\to0$. Continuing by applying Taylor's formula on the far most term, $$\cot(x)=\left(1-\dfrac{x^2}{2}+O(x^4)\right)\frac{1}{x}\left(1+\frac{x^2}{6}+O(x^4)\right)=\frac{1}{x}-\frac{x}{3}+O(x^3)$$ as $x\to0$. It follows that, $$\lim_{x\to0} x\cot(6x)=\lim_{x\to0}\frac{1}{6}-2x^2+O(x^3)=\frac{1}{6}.$$ It is definitely more standard to use $\frac{\sin x}{x}=1$ as $x\to 0$ though.