3
$\begingroup$

How do you compute the following limit without using the l'Hopital rule? If you were allowed to use it, it becomes easy and the result is $\sqrt{3}\over 3$ but without it, I am not sure how to proceed. $$\lim_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$$

$\endgroup$
4
  • 1
    $\begingroup$ i would set $t=x-\frac{\pi}{3}$ $\endgroup$ – Dr. Sonnhard Graubner Dec 14 '15 at 18:05
  • $\begingroup$ I guess you can use the Taylor expansions of sine and cosine. $\endgroup$ – This Is Me Dec 14 '15 at 18:05
  • $\begingroup$ Unnecessary, if you substitute and use the formula for cos(x+y), everything cancels. $\endgroup$ – orion Dec 14 '15 at 18:06
  • $\begingroup$ Can't we use L'Hopital Rule? $\endgroup$ – user297008 Dec 14 '15 at 18:34
1
$\begingroup$

Let $y=x-\pi/3$. Then, the limit of interest becomes

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

$\endgroup$
0
1
$\begingroup$

$\frac {\sin x\cos \frac {\pi}{3}- \cos x\sin\frac {\pi}{3}}{1+2\cos x}\\ \frac {\frac 12\sin x- \frac {\sqrt 3}{2} \cos x}{1-2\cos x}\\ \frac {\sin x- \sqrt 3\cos x}{2-4\cos x}\\ \frac {\sin^2 x- 3\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\ \frac {\sin^2 x + \cos^2 x- 4\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\ \frac {1- 4\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\ \frac {(1- 2\cos x)(1+2\cos x)}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\ \frac {(1+2\cos x)}{2(\sin x + \sqrt 3 \cos x)}$

And now we can plug $x = \frac {\pi}{3}$

$\frac {2}{2(\frac {\sqrt 3}{2} + \frac {\sqrt 3}{2})}$

$\frac {1}{\sqrt 3}$

$\endgroup$
0
$\begingroup$

Proceed as follows:

$$\lim_{x\to\frac{\pi}{3}} \frac{\sin(x-\pi/3)}{1-2\cos x}$$ $$=\lim_{t\to0} \frac{\sin t}{1-2\cos (t+\frac{\pi}{3})}$$ $$=\lim_{t\to0} \frac{\sin t}{1-2\cos t\cos \frac{\pi}{3}+2\sin t \sin\frac{\pi}{3}}$$ $$=\lim_{t\to0} \frac{\sin t}{(1-\cos t)+2\sin t \sin\frac{\pi}{3}}=\frac{1}{2\sin\frac{\pi}{3}}=\frac{1}{\sqrt3}$$

Maybe a little bit of reasoning: $1-\cos t\approx t^2/2$ for small $t$, which is insignificant compared to $\sin t$. If unsure, maybe use $1-\cos t = 2\sin^2 \frac{t}{2}$ and put everything else into half angles to. In that case, you get

$$=\lim_{t\to0} \frac{2\sin \frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}+4\sin\frac{t}{2}\cos\frac{t}{2} \sin\frac{\pi}{3}}=\lim_{t\to0} \frac{2\cos\frac{t}{2}}{2\sin\frac{t}{2}+4\cos\frac{t}{2} \sin\frac{\pi}{3}}=\frac{2}{4\sin\frac{\pi}{3}}=\frac{1}{\sqrt3}$$

$\endgroup$
0

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.