Question

The value of $li{m}_{x\to 0}\frac{cos\left(sinx\right)-cosx}{x^4}$ is equal to

• A. $1/5$
• B. $1/6$
• C. $1/4$
• D. $1/2$

Answer 1

The correct option is B $1/6$

Given,

${lim}_{x\to 0}\left(\frac{\cos \left(\sin \left(x\right)\right)-\cos \left(x\right)}{x^4}\right)$

apply L-Hospital's rule

$={lim}_{x\to 0}\left(\frac{-\sin \left(\sin \left(x\right)\right)\cos \left(x\right)+\sin \left(x\right)}{4x^3}\right)$

again apply L-Hospital's rule

$={lim}_{x\to 0}\left(\frac{-{\cos }^2\left(x\right)\cos \left(\sin \left(x\right)\right)+\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos \left(x\right)}{12x^2}\right)$

once again apply L-Hospital's rule

$={lim}_{x\to 0}\left(\frac{\frac{3\sin \left(2x\right)\cos \left(\sin \left(x\right)\right)+2{\cos }^3\left(x\right)\sin \left(\sin \left(x\right)\right)+2\cos \left(x\right)\sin \left(\sin \left(x\right)\right)-2\sin \left(x\right)}{2}}{24x}\right)$

upon simplification, we get,

$={lim}_{x\to 0}\left(\frac{3\sin \left(2x\right)\cos \left(\sin \left(x\right)\right)+2{\cos }^3\left(x\right)\sin \left(\sin \left(x\right)\right)+2\cos \left(x\right)\sin \left(\sin \left(x\right)\right)-2\sin \left(x\right)}{48x}\right)$

again applying L-Hospital's rule,

$={lim}_{x\to 0}\left(\frac{3(2\cos \left(2x\right)\cos \left(\sin \left(x\right)\right)-\sin \left(\sin \left(x\right)\right)\cos \left(x\right)\sin \left(2x\right))+2(-3{\cos }^2\left(x\right)\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+{\cos }^4\left(x\right)\cos \left(\sin \left(x\right)\right))+2(-\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+{\cos }^2\left(x\right)\cos \left(\sin \left(x\right)\right))-2\cos \left(x\right)}{48}\right)$

substituting $x=0$

$\frac{3(2\cos (2\cdot 0)\cos \left(\sin \left(0\right)\right)-\sin \left(\sin \left(0\right)\right)\cos \left(0\right)\sin (2\cdot 0))+2(-3{\cos }^2\left(0\right)\sin \left(0\right)\sin \left(\sin \left(0\right)\right)+{\cos }^4\left(0\right)\cos \left(\sin \left(0\right)\right))+2(-\sin \left(0\right)\sin \left(\sin \left(0\right)\right)+{\cos }^2\left(0\right)\cos \left(\sin \left(0\right)\right))-2\cos \left(0\right)}{48}$

we get,

$=\frac{1}{6}$