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Find the limit of: $$\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$$

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  • $\begingroup$ hint: $$\cos(2/x)=2\cos^2(1/x)-1$$ $\endgroup$
    – obataku
    Sep 1, 2013 at 16:45
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    $\begingroup$ What, no "without using L'Hospital's rule"? I'm not even sure this site allows asking about limits of fractions without adding an unexplained requirement that the solution must not involve L'Hospital. $\endgroup$ Sep 1, 2013 at 16:46
  • $\begingroup$ The Maple code $$with(Student[Calculus1]): LimitTutor((cos(1/x)-1)/(cos(2/x)-1), x = infinity); $$ finds it step by step. $\endgroup$
    – user64494
    Sep 1, 2013 at 17:53

9 Answers 9

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Let $h=\frac{1}{x}$. We want to find $$\lim_{h\to 0^+} \frac{\cos h-1}{\cos 2h-1}.$$ From the identity $\cos 2h=2\cos^2h-1$, we see that we want $$\lim_{h\to 0^+} \frac{\cos h-1}{2\cos^2 h-2}.$$ But $2(\cos^2 h-1)=2(\cos h-1)(\cos h+1)$, so we want $$\lim_{h\to 0^+} \frac{1}{2(\cos h+1)}.$$ This limit is $\dfrac{1}{4}$.

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Since $\cos t \sim 1-t^2/2$ from Taylor series, then $$ \lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}= \lim_{x\to\infty}{\frac{\frac{1}{2x^2}}{\frac{4}{2x^2}}}=1/4 $$

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    $\begingroup$ How did you get your first equality? If you're using Taylor series perhaps it is a good idea to hint about it... $\endgroup$
    – DonAntonio
    Sep 1, 2013 at 16:49
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    $\begingroup$ @DonAntonio: Thank you, I believed it well known. $\endgroup$ Sep 1, 2013 at 16:54
  • $\begingroup$ it is @BorisNovikov but not for someone being introduced to calculus for the first time $\endgroup$
    – obataku
    Sep 1, 2013 at 16:58
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    $\begingroup$ This is a great question for teachers. (Like me). Solve this problem in at least 5 different ways!!! I am going to use this one for my students. Thanks for the post (+1) I like these kind of problems that have some many ways of how to solve them $\endgroup$
    – imranfat
    Sep 1, 2013 at 17:41
  • $\begingroup$ @imranfat: Thank you. $\endgroup$ Sep 1, 2013 at 17:44
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Let's define: $$h=\frac{1}{x}$$ Then: $$h\to0$$ So we will rewrite the limit as: $$\lim_{h\to0}{\frac{\cos(h)-1}{\cos(2h)-1}}=\lim_{h\to0}{\frac{-\sin(h)}{-2\sin(2h)}}=\lim_{h\to0}{\frac{h}{2\cdot2h}}=\frac{1}{4}$$

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What about a little l'Hospital?

$$\lim_{x\to\infty}\frac{\cos\frac1x-1}{\cos\frac2x-1}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\frac1{x^2}\sin\frac1x}{\frac2{x^2}\sin\frac2x}\stackrel{\text{l'H}}=\frac12\lim_{x\to\infty}\frac{-\frac1{x^2}\cos\frac1x}{-\frac2{x^2}\cos\frac2x}=\frac12\cdot\frac12\cdot\frac11=\frac14$$

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$$\lim_{x \to \infty} \frac{\cos \frac1x - 1}{\cos \frac2x - 1} = \lim_{x \to \infty} \frac{-2\sin^2 \frac{1}{2x}}{-2\sin^2 \frac{1}{x}} = \lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{\sin^2 \frac{1}{x}} = \lim_{x \to \infty} \frac14 \frac{\sin^2 \frac{1}{2x}}{\frac{1}{(2x)^2}} \frac{\frac{1}{x^2}}{\sin^2 \frac{1}{x}} = \frac14.$$

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Rewriting $\cos(2/x)=2\cos^2(1/x)-1$ we have:$$\begin{align*}\lim_{x\to\infty}\frac{\cos(1/x)-1}{\cos(2/x)-1}&=\lim_{x\to\infty}\frac{\cos(1/x)-1}{2\cos^2(1/x)-2}\\&=\frac12\lim_{x\to\infty}\frac{\cos(1/x)-1}{\cos(1/x)^2-1}\\&=\frac12\lim_{x\to\infty}\frac{\cos(1/x)-1}{(\cos(1/x)+1)(\cos(1/x)-1)}\\&=\frac12\lim_{x\to\infty}\frac1{\cos(1/x)+1}\\&=\frac12\cdot\frac12\\&=\frac14\end{align*}$$

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As an alternative to NightRa's answer

$$\mathop {\lim}\limits_{h \to 0} \frac{-\sin h}{-2 \sin {2h}}=\mathop{\lim}\limits_{h\to 0}\frac{-\sin h}{-4 \sin h \cos h}=\mathop{\lim}\limits_{h\to 0}\frac{1}{4\cos h}=\frac{1}{4}.$$

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Another option, let $u=exp(i/2x).$ Then, $\cos(1/x)-1=(u^2+1/u^2)/2-1=(u-1/u)^2/2$ and $\cos(2/x)-1=(u^4+1/u^4)/2-1=(u^2-1/u^2)^2/2$.

Thus: $$\frac{\cos(1/x)-1}{\cos(2/x)-1} =\left(\frac{u-1/u}{u^2-1/u^2}\right)^2 =\left(\frac{1}{u+1/u}\right)^2. $$

Since $u$ tends to $1$ as $x$ goes to infinity, you can conclude that the desired limit is $1/4$.

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$$\lim_{x \to \infty} \frac{\cos \frac1x - 1}{\cos \frac2x - 1} = \lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{\sin^2 \frac{1}{x}} =\lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{(\frac{1}{2x})^2}\frac{(\frac{1}{x})^2}{(\sin^2 \frac{1}{x})}\frac{1}{4}=1*1* \frac14=\frac14.$$

note that $$\lim_{x \to \infty} \frac{\sin \frac1x }{\frac1x}=1$$

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