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$$\lim_{x \to \pi/ 6} \frac{(2\sin x + \cos(6x))^2}{(6x - \pi)\sin(6x)}$$

I would expand with Maclaurin series but $x \to \frac \pi 6$ so I cannot do that. So I evaluated it with l'Hopital rule (result is $-\frac 1{12}$), but is there a better way? I had to differentiate two times and it gets really big and complicated.

user35603
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rubik
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  • And if you change the variable $6x\to t$? Will this help? – Caran-d'Ache Jul 31 '14 at 15:09
  • @Caran-d'Ache: well I tried and expanded with Taylor series. The calculations are not really simpler, I had to expand the series with WolframAlpha: $\lim_{t \to \pi} \frac{(2\sin(\frac t6) + \cos t)^2}{(t-\pi)\sin t} = \lim_{t \to \pi} \frac{\left ( \frac{t - \pi}{2\sqrt{3}} + o(t - \pi) \right )^2}{-(t-\pi)^2 + o((t-\pi)^2))} = \lim_{t \to \pi} - \frac{(t - \pi)^2}{12(t - \pi)^2} = -\frac1{12}$ – rubik Jul 31 '14 at 15:24
  • And now change $t-\pi\to z$? – Caran-d'Ache Jul 31 '14 at 15:36
  • Only after typing my answer did I realise that I might have misinterpreted what you mean with 'better'. I interpreted it as more elementary. Am I incorrect? – Git Gud Jul 31 '14 at 17:16
  • @GitGud: Well I was certainly unclear. Since a direct application of L'Hospital leads to a lot of calculations, by 'better' I meant a more elegant, direct way. Usually in my experience that can be achieved through the Maclaurin expansion, but I found that in this case a lot of calculations were necessary as well. So I was looking at another alternative. – rubik Jul 31 '14 at 17:27
  • @rubik Got it. So I take it some of the answers below are already good enough. – Git Gud Jul 31 '14 at 17:47

4 Answers4

2

You can shift the argument with $x=y+\frac\pi6$:

$$\lim_{y\to0}\frac{(2\sin(y+\frac\pi6)+\cos(6y+\pi))^2}{6y\sin(6y+\pi)}=\lim_{y\to0}\frac{(\sqrt3\sin(y)+\cos(y)-\cos(6y))^2}{-6y\sin(6y)}.$$

Now it is enough to see that Taylor expansion will give $(\sqrt3y+o(y))^2=3y^2+o(y^2)$ in the numerator and $-36y^2+o(y^2)$ in the denominator, and the limit is $$-\frac3{36}.$$

2

An elementary way is the following.

Consider the functions of real variable $f,g$ defined by $f(x)=\sin(6x)$ and $g(x)=2\sin(x)+\cos(6x)$, for all $x\in \mathbb R$. Note that $f(\pi /6)=0=g(\pi /6)$.

Now $$\begin{align} \lim \limits_{x\to \pi /6}\left(\frac{(2\sin x + \cos(6x))^2}{(6x - \pi)\sin(6x)}\right)&=\lim \limits_{x\to \pi /6}\left[\left(\dfrac{2\sin(x)+\cos(6x)}{6x-\pi}\right)^2\dfrac{6x-\pi}{\sin(6x)}\right]\\ &=\left[\lim \limits_{x\to \pi /6}\left(\dfrac{2\sin(x)+\cos(6x)}{6x-\pi}\right)\right]^2\left[\lim \limits_{x\to \pi /6}\left(\dfrac{\sin(6x)}{6x-\pi}\right)\right]^{-1}\\ &=\dfrac 1 {6^2}\left[\lim \limits_{x\to \pi /6}\left(\dfrac{2\sin(x)+\cos(6x)}{x-\frac \pi 6}\right)\right]^2\left(\dfrac 1 6 \right)^{-1}\left[\lim \limits_{x\to \pi /6}\left(\dfrac{\sin(6x)}{x-\frac \pi 6}\right)\right]^{-1}\\ &=\dfrac 1 6 \left[\lim \limits_{x\to \pi/ 6}\left(\dfrac{g(x)-g(\pi /6)}{x-\frac \pi 6}\right)\right]^2\left[\lim \limits_{x\to \pi /6}\left(\dfrac{f(x)-f(\pi /6)}{x-\frac \pi 6}\right)\right]^{-1}\\ &=\dfrac 1 6\left(g'(\pi/ 6)\right)^2\left(f'(\pi /6)\right)^{-1}\\ &=\dfrac 1 6(2\cos(\pi /6)-6\sin(\pi))^2(6\cos(\pi))^{-1}\\ &=\dfrac 1 6\sqrt 3^2\dfrac 1 6 (-1)\\ &=-\dfrac 3 {36}\\ &=-\dfrac 1{12}. \end{align}$$

Git Gud
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1

Write the expression as

$$\left( \frac{2\sin x+\cos 6x}{6x-\pi} \right)^2 \left(\frac{6x -\pi}{\sin 6x}\right)$$

now by L'hopital,

$$\frac{2\sin x+\cos 6x}{6x-\pi} \rightarrow \frac{1}{2\sqrt{3}}$$

$$\frac{6x -\pi}{\sin 6x}\rightarrow -1$$

putting it together you get $-\frac{1}{12}$

Rene Schipperus
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0

1) To simplify, set $6x - \pi = v$,

2) use the angle formulas, i.e. write $\sin x = \sin (\pi -x)$ and $\cos x =-\cos(\pi-x)$.

The first term that easily comes out after multiplying by $v$ is $-\frac{v}{ \sin v} = 1$. Can you do the rest without Taylor series expansion/L'Hospital's?

Alex
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