# Limit of $\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$

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

• hint: $$\cos(2/x)=2\cos^2(1/x)-1$$ Sep 1, 2013 at 16:45
• 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. Sep 1, 2013 at 16:46
• The Maple code $$with(Student[Calculus1]): LimitTutor((cos(1/x)-1)/(cos(2/x)-1), x = infinity);$$ finds it step by step. Sep 1, 2013 at 17:53

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 $\dfrac{1}{4}$.

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

• How did you get your first equality? If you're using Taylor series perhaps it is a good idea to hint about it... Sep 1, 2013 at 16:49
• @DonAntonio: Thank you, I believed it well known. Sep 1, 2013 at 16:54
• it is @BorisNovikov but not for someone being introduced to calculus for the first time Sep 1, 2013 at 16:58
• 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 Sep 1, 2013 at 17:41
• @imranfat: Thank you. Sep 1, 2013 at 17:44

Let's define: $$h=\frac{1}{x}$$ Then: $$h\to0$$ So we will rewrite the limit as: $$\lim_{h\to0}{\frac{\cos(h)-1}{\cos(2h)-1}}=\lim_{h\to0}{\frac{-\sin(h)}{-2\sin(2h)}}=\lim_{h\to0}{\frac{h}{2\cdot2h}}=\frac{1}{4}$$

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

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

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*}

As an alternative to NightRa's answer

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

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 $1/4$.

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

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