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

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

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 $$-\lim_{x \to 0}\frac{\tan(2x)}{x}\frac{\tan(4x)}{x}\frac{\tan(6x)}{x} = -2\cdot 4 \cdot 6 = -48$$

• perfect answer +1 for the same. – Paramanand Singh Jun 24 '16 at 9:31
• Yes if $A+B+C=n\pi$ $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$ here $n=0$ – lab bhattacharjee Jun 24 '16 at 9:37

Hint:

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

You should find $-48$.

HINT:

$$\tan2x-(\tan6x-\tan4x)=\sin2x\left(\dfrac1{\cos2x}-\dfrac1{\cos4x\cos6x}\right)$$

$$=\dfrac{\sin2x(2\cos4x\cos6x-2\cos2x)}{2\cos2x\cos4x\cos6x}$$

Now, $$2\cos4x\cos6x-2\cos2x=\cos2x+\cos10x-2\cos2x$$ $$=\cos10x-\cos2x=-2\sin6x\sin4x$$

Now use $\lim_{h\to0}\dfrac{\sin h}h=1$

$$\underset{x\to 0}{\mathop{\lim }}\,\frac{(\tan 2x-2x)+(\tan 4x-4x)+(6x-\tan 6x)}{{{x}^{3}}}=\frac{1}{3}\underset{x\to 0}{\mathop{\lim }}\,\frac{8{{x}^{3}}+64{{x}^{3}}-216{{x}^{3}}}{{{x}^{3}}}=-48$$