$\displaystyle
\lim_{x \to \pi/3}\frac{\tan^3(x) -3\tan(x)}{\cos(x+\pi/6)}
$
help please
Take the derievative of the top and bottom.
In other words Apply L' Hospitals Rule, it becomes and elementary problem.
$\displaystyle
3\lim_{x \to \pi/3}\frac{\tan^{2}(x)sec^{2} -3\sec^{2}x}{\-sin(x+\pi/6)}
$
Evaluate it and ta-da.
The answer is negative 12 i believe but don't trust me, do it yourselve.
$\displaystyle \lim_{x\to \frac{\pi}{3}}\frac{tan^{3}(x)-3tan(x)}{cos(x+\frac{\pi}{6})}$
Have you learned L'Hopital's rule yet?.
Differentiate top and bottom:
$\displaystyle \lim_{x\to \frac{\pi}{3}}\frac{-3(tan^{4}(x)-1)}{sin(x+\frac{\pi}{6})}$
Now, just plug in x=Pi/3 and you got it. You should get -24