Question

Find the value of $\underset{x\to 0}{lim}\frac{3\tan 3x-4\tan 2x-\tan x}{x^2\tan x}$

Answer 1

$\underset{x\to 0}{lim}\frac{3\tan 3x-4\tan 2x-\tan x}{x^2\tan x}$
$=\underset{x\to 0}{lim}\frac{\frac{3(3\tan x-{\tan }^{3}x)}{1-3{\tan }^{2}x}-\frac{8\tan x}{1-{\tan }^{2}x}-\tan x}{x^2\tan x}$
$=\underset{x\to 0}{lim}\frac{1}{x^2}[3(\frac{1}{3}+\frac{8/3}{1-3{\tan }^{2}x})-1-\frac{8}{1-{\tan }^{2}x}]$
$=\underset{x\to 0}{lim}\frac{8}{x^2}\left[\frac{2{\tan }^{2}x}{(1-{\tan }^{2}x)(1-3{\tan }^{2}x)}\right]=16$