How do you solve this?
$\displaystyle \lim_{x\to\frac{\pi}{3}} (\sqrt{3} \sin x -2x)$
Direct Substitution...
That's all. There's no need for any advanced techniques here.
I'll do it... Here's a step by step of all the relevant theorems of limits that apply.
1. The limit of a sum is the sum of the limits.
$\displaystyle \lim_{x\to\frac{\pi}{3}}(\sqrt{3}\sin x-2x)=\lim_{x\to\frac{\pi}{3}}\sqrt{3}\sin{x}-\lim_{x\to\frac{\pi}{3}}2x$
2. The limit of a product is the product of the limits.
$\displaystyle =\lim_{x\to\frac{\pi}{3}}\sqrt{3}\cdot\lim_{x\to\f rac{\pi}{3}}\sin{x}-\lim_{x\to\frac{\pi}{3}}2\cdot\lim_{x\to\frac{\pi} {3}}x$
4. The limit of a constant is that constant
$\displaystyle =\sqrt{3}\lim_{x\to\frac{\pi}{3}}\sin{x}-2\lim_{x\to\frac{\pi}{3}}x$
5. The ovious limit
$\displaystyle =\sqrt{3}\lim_{x\to\frac{\pi}{3}}\sin{x}-2\frac{\pi}{3}$
6. The limit of transcendentals
$\displaystyle =\sqrt{3}\frac{\sqrt{3}}{2}-2\frac{\pi}{3}=\frac{3}{2}-\frac{2\pi}{3}$
You can see that there is no need for the squeeze theorem.