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

All right, let's go for the Taylor series. We'll be needing terms up to $O(x^6)$.
We'll start by writing $\cos x^{\sin x}$ as $\left(1 + \varepsilon\right)^{\sin x}$, with $\varepsilon \to 0$ as $x \to 0$. The lowest order term in $\sin x$ is x, while the lowest in $\varepsilon$ is $x^2$. To get sufficient detail, we'll therefore need the first two terms in the general binomial formula:
$$(1 + \varepsilon)^{\sin x} = 1 + \varepsilon\sin x + \frac{1}{2!}\varepsilon^2\sin x (\sin x -1) + O(x^7).$$
To the necessary order in $x$, we can write $\varepsilon = -\frac{x^2}{2} + \frac{x^4}{4!} + O(x^6)$ and $\sin x = x - \frac{x^3}{3!} + O(x^5)$. Inserting then gives
$$\begin{align} (1 + \varepsilon)^{\sin x} &= 1 + \left(-\frac{x^2}{2} + \frac{x^4}{4!}\right)\left(x - \frac{x^3}{3!}\right)\\ &\quad + \frac{1}{2} \left(-\frac{x^2}{2} + \frac{x^4}{4!}\right)^2\left(x - \frac{x^3}{3!}\right)\left(x - \frac{x^3}{3!} -1\right) + O(x^7)\\
&= 1 - \frac{x^3}{2} + \frac{x^5}{12} + \frac{x^5}{24} - \frac{x^5}{8} + \frac{x^6}{8} + O(x^7)\\
&= 1 -\frac{x^3}{2} + \frac{x^6}{8} + O(x^7)
\end{align} $$
And for the square root part of the numerator we have 
$$
\sqrt{1 - x^3} = 1 - \frac{x^3}{2} -\frac{x^6}{8} + O(x^9). $$
Combining gives a numerator of $\frac{1}{4}x^6 + O(x^7)$, which divides by the denominator to give $\frac{1}{4} + O(x)$, and so the limit as $x$ tends to zero is indeed $\frac{1}{4}$.
