# Find the limit of this trigonometric expression as x approaches 0

$$\lim_{x\to 0^+} (\cot(x)-\frac{1}{x})(\cot(x)+\frac{1}{x})$$

I have computed the limits

$$\lim_{x\to 0^+} (\cot(x)-\frac{1}{x})=0$$

$$\lim_{x\to 0^+} (\cot(x)+\frac{1}{x})=\infty$$

I can't multiply the limits together.

I rewrote the initial expression:

$$\lim_{x\to 0^+} (\frac{x\cos(x)-\sin(x)}{x\sin(x)})(\frac{x\cos(x)+\sin(x)}{x\sin(x)})$$

$$\lim_{x\to 0^+} \frac{x^{2}\cos^{2}(x)-\sin^{2}(x)}{x^{2}\sin^{2}(x)}$$

$$\lim_{x\to 0^+} \frac{\cos^{2}(x)-1}{\sin^{2}(x)}$$

Using Pythagorean Identity

$$\lim_{x\to 0^+} \frac{-\sin^{2}(x)}{\sin^{2}(x)}=-1$$

I also got the same result using L'Hospital's Rule.

I know the limit is $\frac{-2}{3}$

What did I do wrong?

• There's some algebra problems in the middle... Did you consider $(a+b)(a-b)=a^2-b^2$ right at the beginning? – abiessu Nov 28 '15 at 4:20
• Where did I make the mistake? Yes, I did, but that too gave me -1. I just think there is a very silly error somewhere in there which I can't see. – Daniel Waleniak Nov 28 '15 at 4:29
• Somehow the $x^2$ disappeared going from the second line to the third. They should not have, the "cancellation" is not correct. – André Nicolas Nov 28 '15 at 4:31
• I used the fact that $\lim_{x\to 0} \frac{\sin(x)}{x}=1$ Did I use that incorrectly? – Daniel Waleniak Nov 28 '15 at 4:34
• You cannot (in general) let $x\to 0$ in part of an expression, while leaving some other $x$'s as they were. – André Nicolas Nov 28 '15 at 4:39

A simple way to do it is based on Taylor series; since $$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}+O\left(x^4\right)$$ $$\cot(x)-\frac{1}{x}=-\frac{x}{3}-\frac{x^3}{45}+O\left(x^4\right)$$ $$\cot(x)+\frac{1}{x}=\frac{2}{x}-\frac{x}{3}-\frac{x^3}{45}+O\left(x^4\right)$$ Making the product $$(\cot(x)-\frac{1}{x})\,(\cot(x)+\frac{1}{x})=-\frac{2}{3}+\frac{x^2}{15}+O\left(x^3\right)$$ which shows the limit and how it is approached.

• That's a neat way of approaching it. Thank you – Daniel Waleniak Nov 28 '15 at 4:43
• You are very welcome ! Taylor expansion is simple and solve problems of limits in a glance (infinitely faster than L'Hospital). – Claude Leibovici Nov 28 '15 at 7:37
• I wish we covered more of it. We only did very basic Taylor series in class. – Daniel Waleniak Nov 28 '15 at 7:43
• @DanielWaleniak. If you know Taylor expansions of $\sin(x)$ and $\cos(x)$, long division gives $\tan(x)$ and $\cot(x)$. Simple, isn't it ? – Claude Leibovici Nov 28 '15 at 7:45

It is possible to do not use Taylor series. First write \begin{eqnarray*} \cot ^{2}x-\frac{1}{x^{2}} &=&\left( \frac{x\cos x-\sin x}{x\sin x}\right) \left( \frac{x\cos x+\sin x}{x\sin x}\right) \\ &=&\frac{x^{2}}{\sin ^{2}x}\left( \frac{x\cos x-\sin x}{x^{3}}\right) \left( \frac{x\cos x+\sin x}{x}\right) \\ &=&\left( \frac{x}{\sin x}\right) ^{2}\left( \frac{x\left( \cos x-1\right) +(x-\sin x)}{x^{3}}\right) \left( \cos x+\frac{\sin x}{x}\right) \\ &=&\left( \frac{x}{\sin x}\right) ^{2}\left( \frac{\cos x-1}{x^{2}}+\frac{% x-\sin x}{x^{3}}\right) \left( \cos x+\frac{\sin x}{x}\right) \end{eqnarray*} Basic limits as \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{x}{\sin x} &=&1 \\ \ lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}} &=&-\frac{1}{2} \\ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}} &=&\frac{1}{6} \\ \lim_{x\rightarrow 0}\cos x &=&1 \end{eqnarray*} imply that \begin{eqnarray*} \lim_{x\rightarrow 0}\left( \cot ^{2}x-\frac{1}{x^{2}}\right) &=&\left( \lim_{x\rightarrow 0}\frac{x}{\sin x}\right) ^{2}\left( \lim_{x\rightarrow 0}% \frac{\cos x-1}{x^{2}}+\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}\right) \left( \lim_{x\rightarrow 0}\cos x+\lim_{x\rightarrow 0}\frac{\sin x}{x}% \right) \\ &=&\left( 1\right) ^{2}\left( -\frac{1}{2}+\frac{1}{6}\right) \left( 1+1\right) =-\frac{2}{3}. \end{eqnarray*}

• I think this is a neater way. – Kay K. Nov 28 '15 at 4:59
• This is more elementary, but, as usual, Taylor is easier (in my opinion). – marty cohen Nov 28 '15 at 5:21
• You have a typo - lim(x/sin(x)) should be 1 not zero. – ip6 Nov 28 '15 at 8:20
• @ip6 thank you, i fix it – Idris Nov 28 '15 at 11:54

since $\sin x$ has a very familiar Maclaurin expansion one may note: $$\cot^2 x -\frac1{x^2}= (\sin x)^{-2} - 1 -\frac1{x^2}= x^{-2}(1-\frac{x^2}6+\dots)^{-2} -1 -\frac1{x^2} \\ =x^{-2}(1 +(-2)(-\frac{x^2}6)+O(x^4))) - 1-\frac1{x^2} \\ =-\frac23+O(x^2)$$