Question

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

• A. $1/6$
• B. $-1/3$
• C. $1/2$
• D. $1$

Answer 1

Answer: (B)

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