Finding the limit of a function with a trigonometric exponent I'm sure i'm just missing the trick here, would appreciate some help. 
I tried lhopital and it didn't help, perhaps a trig identity?

 A: Taylor series are the "right" tool to use. Why the right tool? Because the Taylor expansion of $f(x)$ about $x=0$ gives detailed information about the local behaviour of $f(x)$ near $0$.   
However, if you want to use L'Hospital's Rule, proceed as follows. The top and bottom clearly approach $0$.  Note that 
$(\cos x)^{\sin x}=\exp(\sin x \ln(\cos x))$.
If the limit of the ratio of the derivatives exists, so does our limit.  The derivative of the top is
$$\left(\frac{-\sin^2 x}{\cos x} +(\ln(\cos x))(\cos x)\right)   \exp(\sin x \ln(\cos x)) .$$
Things are beginning to look messy!  However, the factor $\exp(\sin x \ln(\cos x))$ approaches $1$, so it is safe, we can forget about it. Quickly we end up wanting the limit of 
$$\frac{\frac{-\sin^2 x}{\cos x} +(\ln(\cos x))(\cos x)}{3x^2}.$$
Two mechanical uses of L'Hospitals's Rule now give the answer. We might fiddle a bit with the expression before differentiating, to make life easier. For example, we can multiply top and bottom by $\cos x$, then forget about the $\cos x$ at the bottom. 
But we can save sweat in various ways, for example by noting that it is a familiar fact that $(\sin^2 x)/(x^2)$ approaches $1$, so now we need only worry about $(\ln(\cos x))/(3x^2)$.
A: compute the taylor sery of $\cos(x)^{\sin(x)}$ by passing throught exp. You get at $0$ 
$$\cos(x)^{\sin(x)}=1-\frac{x^3}{2}+\frac{x^6}{8}+o(x^6).$$
