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Find limit of $$\lim_{x\to 0}\left(\frac{\cos x}{\cos2x}\right)^{x^{-2}}.$$ After some algebra I get $$\lim_{x\to 0}\exp\left(\left(\frac{\cos x-2(\cos x)^2+1}{\cos2x}\right)\cdot\frac{1}{x^2}\right).$$ Maybe I am missing something but I can't continue from this.

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  • $\begingroup$ Expand out the first few terms of $\cos(x)$ and $\cos(2x)$ as a Taylor series, which will give the rate of convergence to one. $\endgroup$
    – Milo Moses
    Mar 1, 2021 at 17:51

6 Answers 6

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Hint:

Let $y = \left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}$. We need to find $\lim_{x\to 0} y$, however, this might get super ugly.

So instead, try to compute $\lim_{x\to 0} \ln(y) = \lim_{x\to 0} \frac{1}{x^2} \ln\left(\frac{\cos(x)}{\cos(2x)}\right) = \lim_{x\to 0} \frac{\ln\left(\frac{\cos(x)}{\cos(2x)}\right) }{x^2}$.

You will have to use L-Hospital twice, and it gets messy algebraically, but very much doable. Then say your limit is $\alpha$, now since $\ln$ is continuous, we know that $\lim_{x \to 0} y = e^{\alpha}$

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I thought it might be instructive to present an approach that relies on straightforward arithmetic and two standard limits and circumvents use of logarithms, L'Hospital's Rule, and Taylor's Theorem. To that end we now proceed.


We begin by writing the term of interest as

$$\begin{align} \left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}&=\left(\frac{2\cos^2(x)-1}{\cos(x)}\right)^{-1/x^2}\\\\ &=\left(1-\underbrace{\left(\frac{1-\cos(x)}{x^2}\right)\left(1+\frac{1+\cos(x)}{\cos(x)}\right)}_{\text{denote this as}\,f(x)}x^2\right)^{-1/x^2}\\\\ &=\left(\left(1-f(x)x^2\right)^{-1/f(x)x^2}\right)^{f(x)} \end{align}$$

Noting that $\lim_{x\to 0}f(x)=3/2$ so that $\lim_{x\to 0}f(x)x^2=0$, and using $\lim_{t\to 0}\left(1-t\right)^{-1/t}=e$, we conclude that

$$\lim_{x\to 0}\left(\frac{\cos(x)}{\cos(2x)}\right)=e^{3/2}$$

NOTE: We tacitly used the "standard limit" $\displaystyle \lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac12$

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  • $\begingroup$ (+) Exactly! Using these theorems is the power of knowledge. But reaching conclusions without using these theorems is an Art. (sorry for my insufficient english) $\endgroup$ Mar 1, 2021 at 18:53
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    $\begingroup$ @lonestudent Your English is more than sufficient. And thank you for the nice note. ;-) $\endgroup$
    – Mark Viola
    Mar 1, 2021 at 18:56
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Rearrange the latter form of the limit that you are getting so that the limit becomes \begin{equation} \lim_{x\to 0}\exp\left(\left(\frac{\cos x-2(\cos x)^2+1}{\cos2x}\right)\cdot\frac{1}{x^2}\right)=\exp\left(\lim_{x\to 0}\dfrac{1-\cos x}{x^2}\times\lim_{x\to0}\dfrac{1+2\cos x}{\cos2x}\right)=\boxed{e^{\frac{3}{2}}} \end{equation}

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First of all $$\lim _{x\to 0}\left(\left(\frac{cosx\:}{cos2x\:}\right)^{\frac{1}{x^2}}\right)=\lim _{x\to 0}\left(e^{\frac{ln\left(\frac{cosx}{cos2x}\right)}{x^2}}\right)=e^{\lim _{x\to 0}\left(ln\left(\frac{cosx}{cos2x}\right)\right)}$$

Now, we will calculate: $$\lim _{x\to 0}\left(\frac{ln\left(\frac{cosx}{cos2x}\right)}{x^2}\right)=\lim \:_{x\to \:0}\left(\frac{ln\left(cosx\right)-ln\left(cos2x\right)}{x^2}\right)=\lim \:\:_{x\to \:\:0}\left(\frac{-tanx+2tan2x}{2x}\right)=\lim \:\:_{x\to \:\:0}\left(\frac{4sec^2x-sec^{2x}}{2}\right)=\frac{3}{2}$$

So the final answer is $e^{\frac{3}{2}}$

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For each $x\ne0$,$$\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}=\exp\left(\frac{\log\left(\frac{\cos(x)}{\cos(2x)}\right)}{x^2}\right)$$and\begin{align}\lim_{x\to0}\frac{\log\left(\frac{\cos(x)}{\cos(2x)}\right)}{x^2}&=\lim_{x\to0}\frac{-\tan(x)+2\tan(2x)}{2x}\\&=\frac12\left(-\lim_{x\to0}\frac{\tan x}x+2\lim_{x\to0}\frac{\tan(2x)}x\right)\\&=\frac32.\end{align}Therefore$$\lim_{x\to0}\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}=e^{3/2}.$$

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Hint

Let $$1+y=1+\dfrac{\cos x-\cos2x}{\cos2x}=1+\dfrac{2\sin\dfrac x2\sin\dfrac{3x}2}{\cos2x}$$

$$(1+y)^{1/x^2}=((1+y)^{1/y})^{y/x^2}$$

Now use $\lim_{y\to0}(1+y)^{1/y}=e$

and $\lim_{h\to0}\dfrac{\sin h}h=1$

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