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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.

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  • $\begingroup$ Are you allowed to use the limit for $\frac{\sin x}x$? $\endgroup$
    – Mark Bennet
    Sep 14, 2014 at 17:55
  • $\begingroup$ yes that is all I am allowed to use. $\endgroup$
    – Alex
    Sep 14, 2014 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, 2014 at 18:06
  • $\begingroup$ Almost definitely not ideal for the third week of Calc 1, but one could add and subtract an appropriate term, then exploit the partial fraction expansion of $\cot z$. $\endgroup$
    – user170231
    Jul 3 at 19:24

4 Answers 4

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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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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,

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

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  • $\begingroup$ Reformatted the answer to make some steps clearer. Kindly revert to previous revision if unacceptable. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Jul 3 at 16:20
  • $\begingroup$ thanks @SarveshRavichandranIyer. IDK mathjax so i just type whatever i want to be displayed in chatgpt to get the appropriate code for it lol $\endgroup$
    – Samarth Saluja
    Jul 3 at 16:34
  • $\begingroup$ Then pick it up from here. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Jul 3 at 16:51
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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.

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