# Calculating $\lim_{x\to \frac {\pi}{6}} \left(\frac {\sin(x- \frac{\pi}{6})}{\frac{{\sqrt3}}{2}-\cos x}\right)$, without using L'Hospital rule

Does anyone have any idea how to transform this to calculate the limit as x approaches $\frac {\pi}{6}$, without using L'Hospital rule?

$$\lim_{x\to \Large{\frac {\pi}{6}}} \left(\frac {\sin(x- \frac{\pi}{6})}{\frac{{\sqrt3}}{2}-\cos x}\right)$$

Let put $t=x-\frac{\pi}{6}$

we will compute

$$\lim_{t\to 0}\frac{\sin(t)}{\frac{\sqrt{3}}{2}-\cos(t+\frac{\pi}{6})}=$$

$$\lim_{t\to0}\frac{\sin(t)}{ \frac{\sqrt{3}}{2}(1- \cos(t))+\frac{\sin(t)}{2} }=$$

$$\lim_{t\to 0}\frac{1}{ \frac{1}{2}+\frac{\sqrt{3}}{2}\frac{t^2(1-\cos(t))}{t^2\sin(t)} }=2.$$

• The first will be the third. – hamam_Abdallah Nov 10 '16 at 21:18
• (1-cos(t)) / (sin(t)) is still just 0/0. Isn't that an issue? – kesaluj Nov 10 '16 at 21:30
• @julasek $\frac{1-\cos(t)}{t^2}$ goes to $\frac{1}{2}$ – hamam_Abdallah Nov 10 '16 at 21:38
• How so? $cos(t)=1$ and $t^2=0$. – kesaluj Nov 10 '16 at 21:40
• $1-cos(t)=2sin^2(\frac{t}{2})$ – hamam_Abdallah Nov 10 '16 at 21:49

Let $f(x) = \sin (x-\pi/6), g(x) = \cos x.$ The expression takes the form

$$-\frac{f(x) - f(\pi/6)}{g(x) - g(\pi/6)} = -\frac{(f(x) - f(\pi/6))/( x-\pi/6)}{(g(x) - g(\pi/6))/(x-\pi/6)}.$$

As $x\to \pi/6,$ the last expression $\to -f'(\pi/6)/g'(\pi/6)$ by the defintion of the derivative. That is easy to calculate, and no L'Hopital was used.

• Lmao, you did not just pull that – Simply Beautiful Art Nov 10 '16 at 21:26
• I don't understand your comment. – zhw. Nov 10 '16 at 22:07
• You basically use L'Hospital's rule, which the OP explicitly said they did not want. – Simply Beautiful Art Nov 10 '16 at 23:01
• Please point out the step where I used L'Hopital. – zhw. Nov 11 '16 at 0:48
• But one is told to not use L'Hopital here, and I didn't. Sorry you cannot understand this elementary point. Everything you've said in these comments is wrong, and I have to say you're pretty rude about it. – zhw. Nov 11 '16 at 16:42

Using trig identities $$\sin 2\theta = 2 \sin\theta \cos\theta\qquad\text{and}\qquad \cos\alpha - \cos\beta = -2 \sin\frac{\alpha - \beta}{2}\sin\frac{\alpha + \beta}{2}$$ ... we can write ...

\begin{align} \sin\left( x - \frac{\pi}{6}\right) &= 2\sin\left(\frac{x}{2} - \frac{\pi}{12}\right) \cos\left(\frac{x}{2} - \frac{\pi}{12}\right) \tag{1} \\[8pt] \frac{\sqrt{3}}{2}-\cos x &= \cos\frac{\pi}{6} - \cos x \\[4pt] &= 2\sin\left(\frac{\pi}{12} - \frac{x}{2}\right) \sin\left(\frac{\pi}{12} + \frac{x}{2}\right) \\[8pt] &= -2\sin\left(\frac{x}{2} - \frac{\pi}{12}\right) \sin\left(\frac{x}{2} + \frac{\pi}{12}\right) \tag{2} \end{align}

... so that, for $\sin\left(\frac{x}{2} -\frac{\pi}{12}\right) \neq 0$ (that is, $x$ in a neighborhood of $\frac{\pi}{6}$ ), ...

$$\frac{\sin\left( x - \dfrac{\pi}{6}\right)}{\dfrac{\sqrt{3}}{2} - \cos x} = - \frac{\cos\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)}{\sin\left(\dfrac{x}{2} + \dfrac{\pi}{12}\right)} \tag{3}$$

Taking the limit on the right-hand side of $(3)$ amounts to substitution.