I found the following limit problem in Differential and Integral Calculus by Piskunov:
$$\lim\limits_{v \to \frac{\pi}{3}}\dfrac{1-2\cos{v}}{\sin{\left(v-\dfrac{\pi}{3}\right)}}$$
I tried searching for a trigonometric identity that could simply this, but couldn't make any work. How can this be solved without the application of L'Hospital's rule?
On
just use Taylor expansions:
$ \cos(x) = \frac{1}{2} - \frac{\sqrt{3} \cdot (x - \frac{\pi}{3})}{2} + o\left( x - \frac{\pi}{3}\right), \ x \to \frac{\pi}{3} $
$ \sin(x) = x - \frac{\pi}{3}, \ o\left( x - \frac{\pi}{3}\right), \ x \to \frac{\pi}{3} $
$ \lim\limits_{x \to \frac{\pi}{3}} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})} = \lim\limits_{x \to \frac{\pi}{3}} \frac{1 - 1 + \sqrt{3} \cdot (x - \frac{\pi}{3}) + o(x - \frac{\pi}{3})}{x - \frac{\pi}{3} + o(x - \frac{\pi}{3})} = \lim\limits_{x \to \frac{\pi}{3}} \sqrt{3} + o(1) = \boxed{\sqrt{3}} $
On
Use Prosthaphaeresis fromula and $\sin2A=2\sin A\cos A$ Formulas,
$$2\cdot\lim_{v\to\pi/3}\dfrac{\cos\dfrac\pi3-\cos v}{\sin\left(v-\dfrac\pi3\right)} =2\cdot\lim_{v\to\pi/3}\dfrac{2\sin\left(\dfrac v2-\dfrac\pi6\right)\sin\left(\dfrac v2+\dfrac\pi6\right)}{2\sin\left(\dfrac v2-\dfrac\pi6\right)\cos\left(\dfrac v2-\dfrac\pi6\right)}$$
Now cancel out $\sin\left(\dfrac v2-\dfrac\pi6\right)$ as $\sin\left(\dfrac v2-\dfrac\pi6\right)\ne0$ as $v\to\dfrac\pi3, v\ne\dfrac\pi3$
You can use the fact that\begin{align}2\cos v&=2\cos\left(\left(v-\frac\pi3\right)+\frac\pi3\right)\\&=\cos\left(v-\frac\pi3\right)-\sqrt3\sin\left(v-\frac\pi3\right).\end{align}It follows from this that$$\frac{1-2\cos v}{\sin\left(v-\frac\pi3\right)}=\frac{1-\cos\left(v-\frac\pi3\right)}{\sin\left(v-\frac\pi3\right)}-\sqrt3$$and therefore all that remains to be done is to compute$$\lim_{t\to0}\frac{1-\cos t}{\sin t}$$and for this you can use the fact that$$1-\cos t=2\sin^2\left(\frac t2\right)\quad\text{and that}\quad\sin t=2\sin\left(\frac t2\right)\cos\left(\frac t2\right).$$