calculus - Evaluating Limit - $(1-(cos x)^{sin x})/(x^3)$

I need to evaluate the following limit using l'Hospital's rule:
$$\lim_{x\to 0}\dfrac{1-(\cos x)^{\sin x}}{x^3}$$

By doing one step, i get
$$\lim_{x\to 0}\dfrac{-(\cos x)^{\sin x}[(\cos x) \ln(\cos x)-\frac{(\sin^2 x)}{\cos x}]}{3x^2}$$

If I did this correctly, I still need to use l'Hospital's rule again, but this seems too complicated for an exam question. Is there another, simpler way of doing this, but by still using L'Hospital's.

Answer

There is a way of solving this using l'Hospital's rule.

$$\lim_{x\to 0}\dfrac{1-(\cos x)^{\sin x}}{x^3}=\lim_{x\to 0}\dfrac{1-(\cos x)^{\sin x}}{\sin^{3}{x}}$$

Apply the rule, we get:

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

$$=\lim_{x\to 0}\dfrac{\ln(\cos x)}{-3x^2}+\frac{1}{3}$$
Apply the rule again:
$$=\lim_{x\to 0}\dfrac{\sin x}{6x\cos x}+\frac{1}{3}$$
$$=\frac{1}{2}$$