Cosine Raised to the Sine?

First Consider $\sin x \ln(\cos x)$ = $(x - \frac{x^3}{3!} + \frac{x^5}{5!} \dots)(\ln ( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \dots))$

= $(x - \frac{x^3}{3!} + \frac{x^5}{5!} \dots)( - \frac{x^2}{2!} + \frac{x^4}{4!} \dots - ( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \dots)^2 \dots)$

= $-\frac{x^3}{2} - \frac{x^7}{80} \dots$

Hence, ${(\cos x)}^{\sin x} = e^{\sin x \ln (\cos x)} = e^{-\frac{x^3}{2} - \frac{x^7}{80} \dots}$

= $1 -\frac{x^3}{2} - \frac{x^7}{80} \dots + (-\frac{x^3}{2} - \frac{x^7}{80} \dots)^2 \dots$ = $1 -\frac{x^3}{2} + \frac{x^{6}}{8} \dots$

Also, using $\sqrt{1-x} = 1 - \frac{x}{2} -\frac{x^2}{8} \dots$ , $\sqrt{1-x^3} = 1 - \frac{x^3}{2} - \frac{x^6}{8} \dots$

$\therefore \displaystyle \lim_{x \to 0} \frac{{(\cos x)}^{(\sin x)} - \sqrt{1 - x^3}}{x^6}$

= $\displaystyle \lim_{x \to 0} \frac{(1 -\frac{x^3}{2} + \frac{x^{6}}{8} \dots) - (1 - \frac{x^3}{2} - \frac{x^6}{8} \dots)}{x^6}$

= $\displaystyle \lim_{x \to 0} \frac{\frac{x^6}{4} \dots}{x^6} = \boxed{\frac{1}{4}}$

I will omit the notation $O(x^7)$ and even " $\cdots$ " to denote higher powers than $x^6$ , and even the limit sign.

$\quad\frac1{x^6}[\cos x^{\sin x}-\sqrt{1-x^3}]$

$=\frac1{x^6}[(1-x^2/2+x^4/24-x^6/720)^{\sin x}-\sqrt{1-x^3}]$

$=\frac1{x^6}[\sum_{k=0}^{\infty}(\sin x\mbox{ choose }k)(-x^2/2+x^4/24-x^6/720)^k-\sum_{k=0}^{\infty}(0.5\mbox{ choose }k)(-x^3)^k]$

$=\frac1{x^6}[1+\sin x(-x^2/2+x^4/24)+(1/2)\sin x(\sin x-1)(x^4/4)-(1-0.5x^3-0.125x^6)]$

$=\frac1{x^6}[1+(x-x^3/6)(-x^2/2+x^4/24)+(1/2)x(x-1)(x^4/4)-(1-0.5x^3-0.125x^6)]$

$=\frac1{x^6}[1+(-x^3/2+x^5/24+x^5/12)+(x^6/8-x^5/8)-(1-0.5x^3-0.125x^6)]$

$=\frac1{x^6}[(1-x^3/2+x^6/8)-(1-0.5x^3-0.125x^6)]$

$=\frac1{x^6}[x^6/4]$

$=\frac14$

I used L Hospital rule,it gives the answer but the procedure is very long and time taking.