Calculate limit using Maclaurin series $$\lim_{x \to 0}\frac{1-(\cos{x})^{\sin{x}}}{x^3}$$
So what I have tried is defining three functions, $f(x)=\cos{x}, g(x) = \sin{x}, h(x) = x^3$
What I did next was do Maclaurin expansion for three functions up until the third order:
$f(x) = 1 - \frac{x^2}{2} + o(x^3)$
$g(x) = x - \frac{x^3}{3} + o(x^3)$
$h(x) = x^3 + o(x^3)$
So when I plug in these values I get:
$$\lim_{x \to 0}\frac{1-(1-\frac{x^2}{2}+o(x^3))^{x-\frac{x^3}{3} + o(x^3)}}{x^3}$$
So after plugging in $x=0$ I get $\frac{0}{0}$ and when I try to apply L'Hôpital's rule I find myself in a situation where I need to calculate the derivative of the denominator several times (which would take too long). I'm positive that I'm correct until this point (I checked on wolframalpha, the result is 1/2).
Does anyone know how to proceed from here? Do I need to calculate the derivative of the denominator several times or is there a smarter way?
 A: I propose an alternative: two steps as follows.
1 - Using MacLaurin Series
We just act on the term $$(\sin(x))^{\cos(x)}$$ by transforming it into an exponential:
$$(\sin(x))^{\cos(x)} = e^{\sin(x)\ln(\cos(x))}$$
Hence it's Taylor-MacLaurin series is $$1 + \sin(x)\ln(\cos(x)) + \ldots$$
Hence we have
$$\lim_{x\to 0} \dfrac{1 - (1 + \sin(x)\ln(\cos(x)))}{ x^3} = \lim_{x\to 0} \dfrac{-\color{red}{\sin(x)}\ln(\cos(x))}{\color{red}{x}\cdot x^2}$$
Notice that the red term goes to $1$ for it is a well known limit.
2 - De L'Hôpital
Now $$\lim_{x\to 0} \dfrac{\ln(\cos(x))}{x^2}$$ can be treated with De L'Hôpital twice, which will give us
$$\lim_{x\to 0} \dfrac{\cos(x)}{2\cos(x) - 2x\sin(x)} = \dfrac{1}{2}$$
A: We can derive the limit using series expansion as follows:
\begin{align*}
\color{blue}{\lim_{x\to 0}}&\color{blue}{\frac{1-\left(\cos x\right)^{\sin x}}{x^3}}\\
&=\lim_{x\to 0}\frac{1-\exp\left((\sin x)\ln\left(\cos x\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(\left(x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)\right)
\ln\left(1-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(\left(x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)\right)
\left(-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(-\frac{x^3}{2}+\mathcal{O}\left(x^4\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\left(1-\frac{x^3}{2}+\mathcal{O}\left(x^4\right)\right)}{x^3}\\
&\,\,\color{blue}{=\frac{1}{2}}
\end{align*}
A: Let us make the limit more "friendly" by a change of variable, $t:=\sin x$. We now have
$$\lim_{t\to0}\frac{1-(1-t^2)^{t/2}}{\arcsin^3t}=\lim_{t\to0}\frac{1-(1-t^2)^{t/2}}{t^3}.$$
Then by the generalized binomial theorem,
$$\lim_{t\to0}\frac{1-1+\dfrac t2t^2-\dfrac t2\left(\dfrac t2-1\right)\dfrac{t^4}2+\cdots}{t^3}$$ and higher order terms. You can conclude.
