Question

Evaluate:${lim}_{x\to 0}\frac{cos\left(sinx\right)-cosx}{x^4}$

Answer 1

${lim}_{x\to o}\frac{cos\left(sinx\right)-cosx}{x^4}$

Using L Hospital rule

${lim}_{x\to o}\frac{d}{dx}\frac{soc\left(sinx\right)-cosx}{\frac{d}{dx}\left(x^4\right)}{lim}_{x\to o}\frac{sin\left(sinx\right)-cosx+sinx}{4x^3}$

again using L Hospital

${lim}_{x\to 0}\frac{+sinx\cdot sin\left(sinx\right)-co{s}^{2}xcos\left(sinx\right)+cosx}{12x^2}$

Using L Hospital rule

${lim}_{x\to 0}\frac{cosx.sin\left(sinx\right)+sinx.cosxcos\left(sinx\right)-sin2xcos\left(sinx\right)-sinx+co{s}^{3}xsin\left(sinx\right)}{24x}$

${lim}_{x\to 0}\frac{cos[1+co{s}^{2}x]sin\left(sinx\right)-sinxcosxcos\left(sinx\right)-sinx}{24x}$

again using L Hospital

${lim}_{x\to 0}\frac{[-sinx-3co{s}^{2}xsinx]Sin\left(sinx\right)+co{s}^{2}x[1+co{s}^{2}x]cos\left(sinx\right)-cosx-os2xcos\left(sinx\right)+si{n}^{2}xcosxsin\left(sinx\right)}{24}$

applying limit

$\Rightarrow \frac{[0-0]0+2-1-1+0}{24}=0$