Question

Solve the problem:
$\underset{x\to 0}{lim}\frac{\mathrm{tanx}+4tan2x-3tan3x}{x^2\tan x}$

• A. $16$
• B. $-16$
• C. $12$
• D. $-12$

Answer 1

The correct option is B $-16$

${lim}_{x\to 0}\frac{\tan x+4\tan 2x-3\tan 3x}{x^2\tan x}$

$={lim}_{x\to 0}\frac{\tan x+\frac{8\tan x}{1-{\tan }^{2}x}-3\left(\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}\right)}{x^2\tan x}$

$={lim}_{x\to 0}\frac{1+\frac{8}{1-{\tan }^{2}x}-3\left(\frac{3-{\tan }^{2}x}{1-3{\tan }^{2}x}\right)}{x^2}$

$={lim}_{x\to 0}\frac{(1-{\tan }^{2}x)(1-3{\tan }^{2}x)+8(1-3{\tan }^{2}x)-3(3-{\tan }^{2}x)(1-{\tan }^{2}x)}{x^2}$

$={lim}_{x\to 0}\frac{1-3{\tan }^{2}x-{\tan }^{2}x+3{\tan }^{4}x+8-24{\tan }^{2}x-9+12{\tan }^{2}x-{\tan }^{4}x}{x^2}$

$={lim}_{x\to 0}\frac{-16{\tan }^{2}x+2{\tan }^{4}x}{x^2}$

$={lim}_{x\to 0}\frac{-16{\tan }^{2}x}{x^2}+{lim}_{x\to 0}\frac{2{\tan }^{2}x}{x^2}{\tan }^{2}x$

$=-16+0=-16$