Limit of $\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$ Find the limit of:
$$\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$$
 A: 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 $\dfrac{1}{4}$.
A: Since $\cos t \sim 1-t^2/2$ from Taylor series, then
$$
\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}= \lim_{x\to\infty}{\frac{\frac{1}{2x^2}}{\frac{4}{2x^2}}}=1/4
$$
A: Let's define:
$$h=\frac{1}{x}$$
Then:
$$h\to0$$
So we will rewrite the limit as:
$$\lim_{h\to0}{\frac{\cos(h)-1}{\cos(2h)-1}}=\lim_{h\to0}{\frac{-\sin(h)}{-2\sin(2h)}}=\lim_{h\to0}{\frac{h}{2\cdot2h}}=\frac{1}{4}$$
A: $$\lim_{x \to \infty} \frac{\cos \frac1x - 1}{\cos \frac2x - 1} = \lim_{x \to \infty} \frac{-2\sin^2 \frac{1}{2x}}{-2\sin^2 \frac{1}{x}} = \lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{\sin^2 \frac{1}{x}} = \lim_{x \to \infty} \frac14 \frac{\sin^2 \frac{1}{2x}}{\frac{1}{(2x)^2}} \frac{\frac{1}{x^2}}{\sin^2 \frac{1}{x}} = \frac14.$$
A: What about a little l'Hospital?
$$\lim_{x\to\infty}\frac{\cos\frac1x-1}{\cos\frac2x-1}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\frac1{x^2}\sin\frac1x}{\frac2{x^2}\sin\frac2x}\stackrel{\text{l'H}}=\frac12\lim_{x\to\infty}\frac{-\frac1{x^2}\cos\frac1x}{-\frac2{x^2}\cos\frac2x}=\frac12\cdot\frac12\cdot\frac11=\frac14$$
A: Rewriting $\cos(2/x)=2\cos^2(1/x)-1$ we have:$$\begin{align*}\lim_{x\to\infty}\frac{\cos(1/x)-1}{\cos(2/x)-1}&=\lim_{x\to\infty}\frac{\cos(1/x)-1}{2\cos^2(1/x)-2}\\&=\frac12\lim_{x\to\infty}\frac{\cos(1/x)-1}{\cos(1/x)^2-1}\\&=\frac12\lim_{x\to\infty}\frac{\cos(1/x)-1}{(\cos(1/x)+1)(\cos(1/x)-1)}\\&=\frac12\lim_{x\to\infty}\frac1{\cos(1/x)+1}\\&=\frac12\cdot\frac12\\&=\frac14\end{align*}$$
A: As an alternative to NightRa's answer
$$\mathop {\lim}\limits_{h \to 0} \frac{-\sin h}{-2 \sin {2h}}=\mathop{\lim}\limits_{h\to 0}\frac{-\sin h}{-4 \sin h \cos h}=\mathop{\lim}\limits_{h\to 0}\frac{1}{4\cos h}=\frac{1}{4}.$$
A: 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 $1/4$.
A: $$\lim_{x \to \infty} \frac{\cos \frac1x - 1}{\cos \frac2x - 1} = \lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{\sin^2 \frac{1}{x}} =\lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{(\frac{1}{2x})^2}\frac{(\frac{1}{x})^2}{(\sin^2 \frac{1}{x})}\frac{1}{4}=1*1* \frac14=\frac14.$$
note that $$\lim_{x \to \infty} \frac{\sin \frac1x }{\frac1x}=1$$
