<s>[INST] 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!  
 [/INST] 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 $\frac{1}{4}$.
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