<s>[INST] Find $\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$ without L'Hopital's rule 
Find the following limit
  $$\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$$
  without using L'Hopital's rule.

I tried to solve this using fundamental limits such as $\lim_{x\to0}\left(1+x\right)^\frac1x=e,\lim_{x\to0}x^x=1$ and equivalent infinitesimals at $x\to0$ such as $x\sim\sin x,a^x\sim1+x\ln a,(1+x)^a\sim1+ax$. This is what I did so far:
$$\begin{align}\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)&=\lim_{x\to0}\frac{\left(\frac{3-(1-\cos x)}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(\frac{3-\frac12x^2}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(1-\frac16x^2\right)^x-1}{x^3}\end{align}$$
I used fact that $x\sim\sin x$ at $x\to0$. After that, I tried to simplify $\left(1-\frac16x^2\right)^x$. Problem is because this is not indeterminate, so I cannot use infinitesimals here. Can this be solved algebraically using fundamental limits, or I need different approach?
 [/INST] Consider first $$A=\left(\frac{2+\cos (x)}{3}\right)^x$$ $$\log(A)=x\log\left(\frac{2+\cos (x)}{3}\right)$$ Now, Taylor series for $\cos(x)$ and then for $\log(1+y)$$$\log(A)=x\log\left(\frac{2+\cos (x)}{3}\right)=x\log\left(\frac{2+1-\frac{x^2}{2}+\cdots}{3}\right)$$ $$\log(A)=x\log\left(\frac{3-\frac{x^2}{2}+\cdots}{3}\right)=x\log\left(1-\frac{x^2}{6}+\cdots\right)$$ $$\log(A)=x\left(-\frac{x^2}{6}+\cdots \right)=-\frac{x^3}{6}+\cdots$$ So $$A\approx e^{-x^3/6+\cdots}$$ Now using Taylor for $e^y$ $$B=\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)\approx\frac1{x^3}\left( e^{-x^3/6}-1\right)=\frac1{x^3}\left(1- \frac {x^3} {6}-1\right)= -\frac {1} {6}$$
If you use one more term in each expansion, you would find $$B=-\frac{1}{6}+\frac{x^3}{72}+\cdots$$ Plotting the curves for $-1<x<1$, you could be amazed to see to see how close is the curve defined by the original expression and the last approximation; they fiffer by less than $0.0002$.
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