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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?

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4 Answers 4

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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.

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  • $\begingroup$ Oh, I have totally forgot to use facts that $\ln(1+x)\sim x$ and $\ln(1-x)\sim-x$ at $x\to0$, this simply solves the problem. However, I solved limit, but I still cannot understand why you think my second expression is wrong? What do you exactly mean? $\endgroup$
    – user164524
    Oct 25, 2015 at 2:55
  • $\begingroup$ Sorry, 2nd expression is fine. $\endgroup$
    – zhw.
    Oct 25, 2015 at 2:58
  • $\begingroup$ This time, perhaps it was I who answered the "headline," but not necessarily the intended question. $\endgroup$
    – Mark Viola
    Oct 25, 2015 at 5:59
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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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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Oct 26, 2015 at 0:34
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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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  • $\begingroup$ Hello my friend! I hope you are well. I posted an answer that relies on standard inequalities only - those that we can establish without calculus - and the squeeze theorem. It's extra work, but I think interesting. $\endgroup$
    – Mark Viola
    Oct 25, 2015 at 6:02
  • $\begingroup$ Hi, my friend ! I saw you good answer, for sure. I just wanted to give another approach. I suppose that you know my love for Taylor. Our affair started almost 60 years ago !! Cheers :-) $\endgroup$ Oct 25, 2015 at 6:36
  • $\begingroup$ Yes. Actually, you posted your solid answer before I posted mine. And I also prefer asymptotics. I wanted to show that there is yet another approach that is quite elementary in that it does not rely on differential or integral calculus. Rather, only the squeeze theorem and standard inequalities. $\endgroup$
    – Mark Viola
    Oct 25, 2015 at 15:13
  • $\begingroup$ Have a nice day, my friend ! $\endgroup$ Oct 25, 2015 at 15:16
  • $\begingroup$ And you have a nice day also! $\endgroup$
    – Mark Viola
    Oct 25, 2015 at 15:17
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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 $$\lim_{x \to 0}\frac{\exp(x) - 1}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2}$$

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  • $\begingroup$ Did you mean $\lim_{x \to 0}\frac{1 - \cos x}{x^2} = \frac{1}{2}$ instead of $\lim_{x \to 0}\frac{1 - \cos x}{2} = \frac{1}{2}$? $\endgroup$
    – user164524
    Oct 26, 2015 at 16:11
  • $\begingroup$ @Mathematician171: Thanks for pointing out. I have fixed the typo. $\endgroup$
    – Paramanand Singh
    Oct 27, 2015 at 2:59

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