# Solving a trigonometric limit $\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$

First off, please excuse my n00bishness I have only just begun learning about algebraic manipulation of limits so this is probably a really dumb or obvious question.

I'm trying to solve the following limit:

$$\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$$ This limit is $0/0$ if evaluated directly, so I tried multiplying by the conjugate of the denominator:

\begin{align} & = \lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{x} + 3)}\\ & = \lim_{x\to\pi/6}\frac{(2\sin{x} - 1)(2\sqrt{3}\cos{x} - 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{3} + 3)} \\ & = \lim_{x\to\pi/6}\frac{2\sin{x} - 1}{2\sqrt{3}\cos{x} + 3}\\ & = \frac{2(1/2) - 1}{2\sqrt{3}\frac{\sqrt{3}}{2} + 3}\\ & = \frac{0}{6}\\ & = 0 \end{align}

But according to WolframAlpha this is incorrect, and the limit should be 1. What have I done wrong?

Also, as I have only just begun I am unfamiliar with L'Hopital's rule.

• Why do you change a sign from the first to the second line? – egreg Jul 12 '14 at 21:59
• Mistake, thanks for catching that. – user3002473 Jul 12 '14 at 21:59
• The error is still there. – egreg Jul 12 '14 at 22:06
• Are you saying $(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3) \neq (2\sin{x} - 1)(2\sqrt{3}\cos{x} - 3)$? I simply flipped the signs from line 1 to line 2 allowing me to cancel out (2\sqrt{3}\cos{x} - 3). – user3002473 Jul 12 '14 at 22:08
• Yes, that's exactly what he's saying. – Code-Guru Jul 12 '14 at 22:08

Hint: $(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{x} + 3)=12\cos^2x-9=3-12\sin^2x$

Now, do you see a way you can use difference of squares to simplify the following expression?$$\lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{3-12\sin^2x}$$

• Oh my gosh I looked at my paper here and I made around 4 or 5 mind-bogglingly dumb mistakes (like saying $(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3) = (2\sin{x} - 1)(2\sqrt{3}\cos{x} - 3)$ along with a few others). Thank you for the help but it turns out my confusion was simply due to lack of attention. – user3002473 Jul 12 '14 at 22:14

You can solve it like this:

$\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}=\frac{2}{2\sqrt{3}} \cdot\lim_{x\to\pi/6}\frac{\frac{1}{2} - \sin{x}}{\cos{x} - \frac{3}{2\sqrt{3}}}=\frac{2}{2\sqrt{3}} \cdot\lim_{x\to\pi/6}\frac{\sin{30^o} - \sin{x}}{\cos{x} - \cos{30^o}}=A$.

Now you can use sum to product trig rules:

$A=\frac{2}{2\sqrt{3}} \cdot\lim_{x\to\pi/6}\frac{2\sin{\frac{30^o-x}{2}}\cos{\frac{{30^0+x}}{2}}}{2\sin{\frac{30^o-x}{2}}\sin{\frac{{30^0+x}}{2}}}=1$ $\blacksquare$

\begin{align} \lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{x} + 3)} \end{align}

You need to multiply out the denominator (and possibly the numerator):

\begin{align} \lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{12\cos^2{x} - 9} \end{align}

As $\displaystyle\sin x-\sin a=\frac{\sin^2x-\sin^2a}{\sin x+\sin a}$

using Prove $\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$, $\displaystyle \sin x-\sin a=\frac{\sin(x+a)\sin(x-a)}{\sin x+\sin a}$

Similarly, $\displaystyle\cos x-\cos a=\frac{\cos^2x-\cos^2a}{\cos x+\cos a}=\frac{1-\sin^2x-(1-\sin^2a)}{\cos x+\cos a}=-\frac{\sin(x+a)\sin(x-a)}{\cos x+\cos a}$

$$\implies\lim_{x\to a}\frac{\sin a-\sin x}{\cos x-\cos a}=\cdots=\lim_{x\to a}\frac{\cos x+\cos a}{\sin x+\sin a}=\cdots$$

Here $\displaystyle a=\frac\pi6$

This method can be safely employed

$$\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$$
$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{\cos x-\cos\frac\pi6}$$
$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{x-\frac\pi6}\cdot\frac1{\lim_{x\to\pi/6}\dfrac{\cos x-\cos\frac\pi6}{x-\frac\pi6}}$$
$$=-\frac2{2\sqrt3}\frac{\dfrac{d(\sin x)}{dx}_{(\text{ at } x=\frac\pi6)}}{\dfrac{d(\cos x)}{dx}_{(\text{ at } x=\frac\pi6)}}$$
$$=\cdots$$
Another way is to use change of variables. Let $y = x - \pi/6$. $$\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3} = \lim_{y \to 0}\frac{1 - 2\sin{(y + \pi/6)}}{2\sqrt{3}\cos{(y+\pi/6)} - 3} = L$$ But, $$1 - 2\sin(y + \pi/6) = (1 - \cos y) - \sqrt{3}\sin y$$ and $$2\sqrt{3}\cos{(y+\pi/6)} - 3 = 2\sqrt{3}\bigg(\frac{\sqrt{3}}{2}\cos y - \frac{1}{2}\sin y\biggr) - 3$$ $$= 3\cos y - \sqrt{3}\sin y - 3 = -3(1 - \cos y) - \sqrt{3}\sin y$$ Thus, $$L = \lim_{y \to 0}\dfrac{\dfrac{1 - \cos y}{y} - \dfrac{\sqrt{3}\sin y}{y}}{-3\dfrac{1 - \cos y}{y} - \sqrt{3}\dfrac{\sin y}{y}} = 1$$ I used the fact that $$\lim_{y \to 0}\frac{1 - \cos y}{y} = 0 \quad \text{and} \quad \lim_{y \to 0}\frac{\sin y}{y} = 1$$