# A trigonometric limit $\lim_{x\to 0}\left( \frac {\cos(\pi x)}{\sin(\pi x)}-\frac{\pi x}{\sin^2 (\pi x)}\right)=0$

The software maple 12 has calculated that:

$\displaystyle \lim_{x\to 0}\Bigg( \frac {\cos(\pi x)}{\sin(\pi x)}\;\;-\;\frac{\pi x}{\sin^2 (\pi x)}\bigg)=0$

How can I prove this equality? I have tried to multiply $\displaystyle \frac{\pi x}{\pi x}$ and use the limit $\displaystyle\frac{\sin(\pi x)}{\pi x}\;=1$ but probably it is the wrong way.

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One possible method for limits like these is to first find a common denominator, then, after combining the fractions, expand the numerator and the denominator as power series. – Antonio Vargas Aug 27 '12 at 20:05
The first thing to do is to replace the annoying $\pi x$ with $t$. Saves typing. – André Nicolas Aug 27 '12 at 20:13

Using l’Hospital’s rule:

\begin{align*} \lim_{x\to 0}\left(\frac{\cos\pi x}{\sin\pi x}-\frac{\pi x}{\sin^2\pi x}\right)&=\lim_{x\to 0}\frac{\sin\pi x\cos\pi x-\pi x}{\sin^2\pi x}\\ &=\lim_{x\to 0}\frac{\sin 2\pi x-2\pi x}{2\sin^2\pi x}\\ &=\lim_{x\to 0}\frac{2\pi\cos2\pi x-2\pi}{4\pi\sin\pi x\cos\pi x}\\ &=\lim_{x\to 0}\frac{\cos2\pi x-1}{\sin2\pi x}\\ &=\lim_{x\to 0}\frac{-2\pi\sin2\pi x}{2\pi\cos\pi x}\\ &=0\;. \end{align*}

Edit: Very silly incorrect computation without l’Hospital’s rule deleted.

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The computations in Without l’Hospital’s rule should be modified. (Already 3 upvotes... Makes one wonder whether people actually read the solutions, or do they simply trust the author's reputation. And just now, the OP accepted this answer! :-)) – Did Aug 27 '12 at 20:19
the second computation is wrong.. – Dubious Aug 27 '12 at 20:22
Brian: To be clear, I know your solution will be all right in a few minutes. My previous comment was more about the general ways along which the site is functioning... – Did Aug 27 '12 at 20:23
Brian: Let me respectfully suggest to save the second computation (you simply forgot a sine in a denominator, otherwise, the method goes through!). – Did Aug 27 '12 at 20:27
@did: I actually made two algebraic errors, and I don’t see a way to make it go through: it essentially boils down to $\lim_{x\to 0}\left(\frac1x-\frac1{\sin x}\right)$, and I don’t see a purely elementary way to show that that’s $0$. (Then again, I may just have ein Brett vorm Kopf.) – Brian M. Scott Aug 27 '12 at 21:12

You can just compute an asymptotic expansion of these functions:

$$\frac{\cos(\pi x)}{\sin (\pi x)} =_{x\to 0} \frac{1}{\pi x}-\frac{\pi x}{3} + \mathcal O(x^3)$$

and

$$\frac{\pi x}{\sin^2 (\pi x)} =_{x\to 0} \frac{1}{\pi x} + \frac{\pi x}{3} + \mathcal O(x^3)$$

So by taking the difference, you get your limit.

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Any hint on how you calculated the limits above? Thanks. – NoChance Aug 27 '12 at 20:13
You divide the Taylor expansion of $\cos(\pi x)$ by the T.e. of $\sin(\pi x)$. You can also take the square of a T.e. and remove terms of higher order and divide them, etc. You can see: en.wikipedia.org/wiki/Taylor_series – Ilies Zidane Aug 27 '12 at 20:19
Thanks for the explanation. – NoChance Aug 27 '12 at 21:04

I would multiply the first by $\frac {\sin (\pi z)}{\sin (\pi z)}$ to put them over a common denominator, then expand numerator and denominator in a Taylor series.

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you have to compute the following limit

$$\displaystyle \lim_{x\to 0}\Bigg( \frac {\cos(\pi x)}{\sin(\pi x)}\;\;-\;\frac{\pi x}{\sin^2 (\pi x)}\bigg)$$

$\displaystyle \frac{cos (\pi x)\cdot sin (\pi x)}{sin^2 (\pi x)}-\frac{\pi x}{sin^2 (\pi x)}=\frac{2cos (\pi x)\cdot sin (\pi x)}{2sin^2 (\pi x)}-\frac{2\pi x}{2sin^2 (\pi x)}=\frac{sin (2\pi x)-2\pi x}{2sin^2 (\pi x)}=_{x \rightarrow 0} =\frac{0}{0}$ and now we can apply the L'Hopital.

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