Calculation of some limit I'm having problems with limits like that $ \lim_{x → 0} \left(\frac{1- \left(\cos x\right)^{\sin x}} {x^3}\right)$. I need to calculate it in pretty rigorous way.
I can use Taylor and write $\cos x^{\sin x}$ as $\left(1+o\left(x^2\right)\right)^{x + o\left(x^2\right)}$ but I have no idea what to do with such expression.
 A: $$(\cos(x))^{\sin(x)}=$$
$$e^{\sin(x)\ln(1-2(\sin(\frac{x}{2})  )^2 )}=$$
$$1+(x+x\epsilon_1(x))\left(-\frac{x^2}{2}+x^2\epsilon_2(x)\right)=$$
$$1-x^3\left(\frac{1}{2}+\epsilon(x)\right).$$
the limit is
$$\frac{1}{2}$$
A: You can compose polynomial expansions. We'll use expansions at order $3$.
By definition, $\;\cos x^{\sin x}=\mathrm e^{\sin x\ln(\cos x)}$.
Now $\;\cos x=1-\dfrac{x^2}2+o(x^3)$, and $\;\ln(1-u)=-u-\dfrac{u^2}2-\dfrac{u^3}3-\dotsm$, so it will be enough to use the expansion of $\;\ln(1-u)$ at order $1$:
$$\ln(\cos x)=-\frac{x^2}2+o(x^3),$$
$$\text{so}\quad\sin x\,\ln(\cos x)=\Bigl(x-\frac{x^3}6+o(x^3)\Bigr)\Bigl(-\frac{x^2}2+o(x^3)\Bigr)=-\frac{x^3}2+o(x^3).$$
Now expand $\;\mathrm e^u$ at order $1$:
$$1-\mathrm e^{\sin x\ln(\cos x)}=1-\Bigl(1-\frac{x^3}2+o(x^3)\Bigr)=\frac{x^3}2+o(x^3)\sim_0\frac{x^3}2,$$
and finally
$$\frac{1-\cos x^{\sin x}}{x^3}\sim_0\frac{\dfrac{x^3}2}{x^3}=\frac12.$$
A: Use the generalized binomial theorem:
$$(1+o(x^2))^{x+o(x^2)}=1+(x+o(x^2))o(x^2)+(x+o(x^2))(x+o(x^2)-1)\frac{(o(x^2))^2}{2!}\cdots$$
