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

Let, $$\text{L}= \displaystyle \lim_{x\to 0}\Bigg( \frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}\Bigg) $$

$\implies \text{L}=\displaystyle \lim_{x \to 0}\dfrac{e^{\sin(x) \times \ln(\cos(x))}-e^\frac {\ln(1-x^3)}{2}}{x^6}$

$\implies \text{L}= \displaystyle \lim_{x \to 0}e^\frac {\ln(1-x^3)}{2} \times \left(\dfrac{e^{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}-1}{x^6}\right)$

$\implies \text{L}=\displaystyle\lim_{x \to 0}\dfrac{e^{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}-1}{x^6}$

$\implies \text{L}=\displaystyle\lim_{x \to 0}\dfrac{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}{x^6}$ $\left[\text{Using} \displaystyle\lim_{x\to 0}\left( \dfrac{e^x - 1}{x} = 1\right)\right]$

Now, since the limit exists, we infer,

$\text{L.H.L.} = \text{R.H.L.} = \text{L}$

$\implies 2\text{L} = \text{L.H.L.} + \text{R.H.L.}$

$=\displaystyle\lim_{x\to 0^-}\dfrac{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}{x^6} + \displaystyle\lim_{x\to 0^+}\dfrac{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}{x^6}$

=$\displaystyle\lim_{x\to 0}\dfrac{\sin(-x) \times \ln(\cos(-x))-\frac {\ln(1+x^3)}{2}}{x^6} + \displaystyle\lim_{x\to 0}\dfrac{\sin(x) \times \ln(\cos(x))-\frac {\ln(1-x^3)}{2}}{x^6}$

$\left[\text{Using} \displaystyle\lim_{x\to 0^-}f(x) = \displaystyle\lim_{x\to 0}f(0-x)\right] $

=$\displaystyle\lim_{x\to 0} \dfrac{\frac{-\ln(1-x^6)}{2}}{x^6}$

=$\dfrac{-(-x^6)}{2x^6}$ $\left[\text{Using} \displaystyle\lim_{x\to 0}\dfrac{\ln(1+x)}{x}=1\right]$

=$\dfrac{1}{2}$

$\implies \boxed{\text{L}=\dfrac{1}{4}}$

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 $\frac{1}{4}$.

If you're given that the limit exists (as the OP stipulates), then you can have all kinds of fun computing what the limit is. To begin with, you can get rid of the square root symbol by multiplying top and bottom by $(\cos x)^{\sin x}+\sqrt{1-x^3}$, which has limit $2$ as $x\to0$, yielding

$$L=\lim_{x\to0}{(\cos x)^{2\sin x}-(1-x^3)\over2x^6}$$

Now use the fact that $\cos x$ is and even function and $\sin x$ is an odd function to get

$$\lim_{x\to0}{(\cos x)^{2\sin x}-(1-x^3)\over2x^6}=\lim_{x\to0}{(\cos x)^{-2\sin x}-(1+x^3)\over2x^6}$$

which allows us to conclude that

$$2L=\lim_{x\to0}{(\cos x)^{2\sin x}-2+(\cos x)^{-2\sin x}\over2x^6}={1\over2}\lim_{x\to0}\left({(\cos x)^{(\sin x)}-(\cos x)^{-(\sin x)}\over x^3}\right)^2$$

Thus it remains to show that

$$\lim_{x\to0}{(\cos x)^{(\sin x)}-(\cos x)^{-(\sin x)}\over x^3}=\lim_{x\to0}{(\cos x)^{2\sin x}-1\over x^3}=\pm1$$

(Note, the simplification of the limit in the last line uses the obvious limit $\lim_{x\to0}(\cos x)^{\sin x}=1^0=1$.) This is fairly easily done in a couple of L'Hopital steps, with simplifications along the way that ultimately boil things down to $\lim_{x\to0}{\sin x\over x}=1$:

$$\begin{align} \lim_{x\to0}{(\cos x)^{2\sin x}-1\over x^3}&=\lim_{x\to0}{2\left(\cos x\ln(\cos x)-{\sin^2x\over\cos x}\right)(\cos x)^{2\sin x}\over 3x^2}\\ &={2\over3}\left(\lim_{x\to0}{\ln(\cos x)\over x^2}-\lim_{x\to0}{\sin^2x\over x^2}\right)\\ &={2\over3}\left(\lim_{x\to0}{{-\sin x\over\cos x}\over2x}-1\right)\\ &={2\over3}\left(-{1\over2}-1\right)=-1 \end{align}$$

The key step where the assumption that the limit exists came into play was when the two versions of the limit, with $x$ and $-x$ were combined into the expression for $2L$, eliminating the $x^3$ in the numerator. Formally, that term would drop out for any odd power of $x$ (or, for that matter, any odd function whatsoever), but clearly the limit doesn't exist for just any such function. So if the assignment were to show that the limit exists as well as to evaluate it, this approach does not do the job.

Added later: I finally read MathGod's answer carefully, and see that our approaches are quite similar. We both make use of the formalism $(\lim_{x\to a}f(x))(\lim_{x\to a}g(x))=\lim_{x\to a}(f(x)g(x))$ to simplify various expressions, and, more crucially, we both use the symmetry between $x$ and $-x$ to get rid of a problematic odd function, leaving an expression for $2L$ that's (relatively) easy to evaluate. The main difference is that MathGod gets rid of the trig function, leaving something that can be dealt with directly, whereas I get rid of an $x^3$, leaving something that can be recognized as a square. Overall, I like MathGod's approach better (now that I understand it), because its simplifications get to the final limit more quickly, without any need for L'Hopital (aside from its implicit use in the special limits for $(e^x-1)/x$ and $(\log(1+x))/x$).