TITLE: Evaluate $\lim_{x\to 0}\frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}$
QUESTION [19 upvotes]: Evaluate
  $$ \displaystyle \lim_{x\to 0}\Bigg( \frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}\Bigg).$$

I tried to use L'Hopital's rule but it got very messy. Moreover I also tried to analyze from graphs, but I was getting the limit $= 0$ by observing it. However, the answer given in my book is $\frac{1}{4}$. Is there any method to do without Taylor series and L' Hopital's rule (like using special limits). We are given that the limit exists. Any help will be appreciated.
Thanks!

REPLY [2 votes]: $$\begin{aligned}
\lim _{x\to 0}\left(\frac{\left(\cos \:\:x\right)^{\sin \:\:x}\:-\:\sqrt{1\:-\:x^3}}{x^6}\right)
\\&=\lim _{x\to 0}\left(\frac{1-\frac{x^3}{2}+\frac{x^6}{8}+o\left(x^6\right)\:-\:\left(1-\frac{x^3}{2}-\frac{x^6}{8}+o\left(x^6\right)\right)}{x^6}\right)
\\&=\color{red}{\frac{1}{4}}
\end{aligned}$$
Solved with Taylor expansion