Find $\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$ without L'Hopital's rule

Question

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?

Answer 1

You are certainly on the right track (although you've used an approximation that needs justification). Now write

$$(1-x^2/6)^x = \exp (x\ln(1-x^2/6)).$$

More approximations: $\ln (1-h) \approx -h$ and $e^h \approx 1 +h $ for $h$ small. What happens if $\approx$ is replaced by $=$ here? That will tell you what the limit is. Now you need to make sure these approximations really work.

Answer 2

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 $0.0002$.

Answer 3

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

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{1}{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)=-\frac16}$$

Answer 4

We can proceed in the following manner \begin{align} L &= \lim_{x \to 0}\frac{1}{x^{3}}\left(\left(\frac{2 + \cos x}{3}\right)^{x} - 1\right)\notag\\ &= \lim_{x \to 0}\frac{1}{x^{3}}\left(\exp\left(x\log\left(\frac{2 + \cos x}{3}\right)\right) - 1\right)\notag\\ &= \lim_{x \to 0}\frac{1}{x^{3}}\cdot x\log\left(\dfrac{2 + \cos x}{3}\right)\cdot\dfrac{\exp\left(x\log\left(\dfrac{2 + \cos x}{3}\right)\right) - 1}{x\log\left(\dfrac{2 + \cos x}{3}\right)}\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\cdot \log\left(\dfrac{2 + \cos x}{3}\right)\cdot 1\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\cdot \log\left(1 + \frac{\cos x - 1}{3}\right)\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\cdot \dfrac{\cos x - 1}{3}\cdot\dfrac{\log\left(1 + \dfrac{\cos x - 1}{3}\right)}{\dfrac{\cos x - 1}{3}}\notag\\ &= -\frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\notag\\ &= -\frac{1}{6}\notag \end{align} We have used the following standard limits $$\lim_{x \to 0}\frac{\exp(x) - 1}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2}$$