<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] Another approach starts with using the half-angle formula for the sine function to establish the identity
$$\frac{2+\cos x}{3}=1-\frac23 \sin^2(x/2)$$
Then, recall the inequalities
$$\frac{-x}{1-x}\le \log(1-x)\le -x \tag 1$$
$$1-x\le e^{-x}\le \frac{1}{1+x} \tag 2$$
and 
$$\frac14 x^2\cos^2 (x/2) \le \sin^2 (x/2)\le \frac14 x^2 \tag 3$$

Note that we can establish $(1)$ and $(2)$ using nothing more than the limit definition $e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n$ of the exponential function, the fact that $\left(1-\frac xn\right)^n$ is a decreasing sequence See This Answer, and Bernoulli's Inequality.  In addition, we can establish $(3)$ by appealing to the geometric interpretation of the sine and cosine functions. 

Then, using $(1)-(3)$, for $x>0$ reveals 
$$\begin{align}
\frac{1}{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)&=\frac{1}{x^3}\left(e^{x\log\left(1-\frac23 \sin^2(x/2)\right)}-1\right)\\\\
&\le-\frac16 \frac{\cos^2(x/2)}{1+\frac16x^3\cos^2(x/2)}
\end{align}$$
and $$\begin{align}
\frac{1}{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)&=\frac{1}{x^3}\left(e^{x\log\left(1-\frac23 \sin^2(x/2)\right)}-1\right)\\\\
&\ge-\frac16 \frac{1}{1-\frac23 \sin^2(x/2)}
\end{align}$$
Therefore, the right-sided limit 
$$\lim_{x\to 0^+}\frac{1}{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)=-\frac16$$
A similar development for $x<0$ shows that the left-sided limit is also $-1/6$.  Therefore, using only (i) standard inequalities that can be established without using calculus and (ii) the squeeze theorem yields
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{1}{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)=-\frac16}$$
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