Q:

What is $\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x + \tan 4x - \tan 6x}}{{{x^3}}}$? .

What is $\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x + \tan 4x - \tan 6x}}{{{x^3}}}$?

A:

Hint:
Use Taylor's formula at order $3$: $\quad\tan u=u+\dfrac{u^3}3+o(u^3)$.
You should find $-48$.

A:

Note that
$$\tan A + \tan B - \tan(A+B)=(1-\tan A \tan B)\tan(A+B)-\tan(A+B) \\=-\tan A \tan B \tan(A+B)$$
So your limit is simply $$-\lim_{x \to 0}\frac{\tan(2x)}{x}\frac{\tan(4x)}{x}\frac{\tan(6x)}{x} = -2\cdot 4 \cdot 6 = -48$$

