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.
 A: You can simplify this a bit by using some standard limits, like $$\lim_{x\to0} \frac{e^x-1}{x} = 1$$ and $$\lim_{x\to0} \frac{\ln(1+x)}{x}=1$$.
Therefore, $$\lim_{x\to0} \frac{1-(\cos x)^{\sin x}}{x^3} = -\lim_{x\to0}\frac{e^{\sin x\ln(\cos x)}-1}{\sin x\ln(\cos x)}\cdot\frac{\sin x\ln(\cos x)}{x^3} = 
-\lim_{x\to0} \frac{\sin x\ln(1+\cos x-1)}{\cos x-1}\cdot \frac{\cos x-1}{x^3} = -\lim_{x\to0} \frac{\sin x\cos x-\sin x}{x^3} = -\lim_{x\to0} \frac{\sin(2x)-2\sin x}{2x^3}$$
And the last limit is indeed a lot more approachable than the initial one.
A: 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:
$$=\lim_{x\to 0}\dfrac{\sin x}{6x\cos x}+\frac{1}{3}$$
$$=\frac{1}{2}$$
A: You can greatly simplify the computations if you remove factors that you know have limit $1$. First, the derivative of $f(x)=(\cos x)^{\sin x}$ is obtained by $\log f(x)=\sin x\log\cos x$, so it is
$$
\frac{f'(x)}{f(x)}=\cos x\log\cos x-\frac{\sin^2x}{\cos x}=
\cos x(\log\cos x-\tan^2x)
$$
so after the first application of l'Hôpital you get
$$
\lim_{x\to0}f(x)\cos x\frac{\tan^2x-\log\cos x}{3x^2}
$$
and you can disregard $f(x)\cos x$, because it has limit $1$. Therefore you need
$$
\lim_{x\to0}\frac{\tan^2x-\log\cos x}{3x^2}=
\lim_{x\to0}\frac{2\tan x(1+\tan^2x)+\tan x}{6x}=
\lim_{x\to0}\frac{3\tan x+2\tan^3x}{6x}=\frac{1}{2}
$$
A different strategy is using Taylor expansions: you can use
$$
(\cos x)^{\sin x}=(\cos^2x)^{(\sin x)/2}=
(1-\sin^2x)^{\sin x/2}
$$
so you can rewrite your limit as
$$
\lim_{x\to0}\frac{1-(1-\sin^2x)^{(\sin x)/2}}{\sin^3x}\frac{\sin^3x}{x^3}
=
\lim_{t\to0}\frac{1-(1-t^2)^{t/2}}{t^3}
$$
because the second fraction has limit $1$ and you can do the substitution $t=\sin x$, which is bijective (and continuous) in a neighborhood of $0$.
Now
$$
(1-t^2)^{t/2}=\exp\left(\frac{t}{2}\log(1-t^2)\right)
=
1+\frac{t}{2}\log(1-t^2)+o(t^3)=1-\frac{t^3}{2}+o(t^3)
$$
and so you have
$$
\lim_{t\to0}\frac{1-1+t^3/2+o(t^3)}{t^3}=\frac{1}{2}
$$
