Question

$\underset{x\to 0}{lim}\frac{\cos \left(\tan x\right)-\cos x}{x^4}$ is equal to

• A. $\frac{1}{6}$
• B. $\frac{-1}{3}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is B $\frac{-1}{3}$

$\cos \left(\tan x\right)-\cos x=2\sin \left(\frac{x+\tan x}{2}\right)\sin \left(\frac{x-\tan x}{2}\right)$
or $\underset{x\to 0}{lim}\frac{\cos \left(\tan x\right)-\cos x}{x^4}=\underset{x\to 0}{lim}\frac{2\sin \left(\frac{x+\tan x}{2}\right)\sin \left(\frac{x-\tan x}{2}\right)}{x^4}$
$=\underset{x\to 0}{lim}\frac{2\sin \left(\frac{x+\tan x}{2}\right)\sin \left(\frac{x-\tan x}{2}\right)}{x^4\left(\frac{x+\tan x}{2}\right)\left(\frac{x-\tan x}{2}\right)}\left(\frac{x^2-{\tan }^{2}x}{4}\right)$
$=\frac{1}{2}\underset{x\to 0}{lim}\frac{x^2-{\tan }^{2}x}{x^4}$
$=\frac{1}{2}\underset{x\to 0}{lim}\frac{x^2-{(x+\frac{x^3}{3}+\frac{2}{15}{x}^5+.....)}^2}{x^4}$
$=\frac{1}{2}\underset{x\to 0}{lim}\frac{1}{x^2}(1-{(1+\frac{x^2}{3}+\frac{2}{15}{x}^4+...)}^2)=-\frac{1}{3}$