3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why do you change a sign from the first to the second line? $\endgroup$ – egreg Jul 12 '14 at 21:59
  • $\begingroup$ Mistake, thanks for catching that. $\endgroup$ – user3002473 Jul 12 '14 at 21:59
  • $\begingroup$ The error is still there. $\endgroup$ – egreg Jul 12 '14 at 22:06
  • $\begingroup$ 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). $\endgroup$ – user3002473 Jul 12 '14 at 22:08
  • $\begingroup$ Yes, that's exactly what he's saying. $\endgroup$ – Code-Guru Jul 12 '14 at 22:08
3
$\begingroup$

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?$$\lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{3-12\sin^2x}$$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user3002473 Jul 12 '14 at 22:14
2
$\begingroup$

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:

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

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$
1
$\begingroup$

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

in How to evaluate this trigonometric limit?

and in Find the limit as $x$ tends towards $\frac{\pi}{4}$

$\endgroup$
  • $\begingroup$ @user3002473, How about this? $\endgroup$ – lab bhattacharjee Jul 14 '14 at 16:56
0
$\begingroup$

$$\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$$

$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{\cos x-\cos\frac\pi6}$$

$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{x-\frac\pi6}\cdot\frac1{\lim_{x\to\pi/6}\dfrac{\cos x-\cos\frac\pi6}{x-\frac\pi6}}$$

$$=-\frac2{2\sqrt3}\frac{\dfrac{d(\sin x)}{dx}_{(\text{ at } x=\frac\pi6)}}{\dfrac{d(\cos x)}{dx}_{(\text{ at } x=\frac\pi6)}}$$

$$=\cdots$$

$\endgroup$
0
$\begingroup$

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 $$ \lim_{y \to 0}\frac{1 - \cos y}{y} = 0 \quad \text{and} \quad \lim_{y \to 0}\frac{\sin y}{y} = 1 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.