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Find $\lim_{t\to 0} \frac{\cos(2t)-1}{\cos(t)-1}$.

Of course, we can't have it in for $0/0$. How would I solve this using limit Laws, please? Is there a way to solve this only using limit laws? I tried asking for help but they used L'hospital rule, but we haven't learnt that yet in class. I'm also not that great with trig Identities but I looked online to find the identity of $\cos(2t)-1$ and the identity of $\cos(t)-1$ to no avail. Am I forgetting a step?

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6 Answers 6

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Only with limit laws and using

  • $\cos 2t = 2\cos^2 t -1$

$$\frac{\cos 2t - 1}{\cos t -1} = 2\frac{\cos^2 t -1}{\cos t - 1}= 2(\cos t + 1)\stackrel{t \to 0}{\longrightarrow}2\cdot 2 = 4$$

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Let $c:=\cos t$ so $\frac{\cos 2t-1}{\cos t-1}=\frac{2c^2-2}{c-1}=2c+2$. As $t\to0$, $c\to1$, so the limit is $2\times1+2=4$.

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$\cos (2t)=1-2\sin ^{2}(t)$ and $\cos t=1-2\sin ^{2} (t/2)$. So $\frac {\cos (2t)-1} {\cos t -1} =\frac {\sin^{2}(t)} {\sin^{2}(t/2)}$. This can be written as $ 4 (\frac {\sin (t)} t)^{2} (\frac {\sin (t/2)} {t/2})^{-2}$. Using the fact that $\frac {sinx}x \to 1$ as $x \to 0$ we see that the limit is $4$.

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It is $\frac {\sin^2(t)}{\sin^2(\frac{t}{2})}=4.\cos^2(\frac{t}{2})$

Now use the fact that $\lim n\to 0, \cos(n)\to1$

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Hint:) Write $\cos(2t) = 2\cos^2t-1$.

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Hint:

$$\lim_{x\to0}\dfrac{\cos2x-1}{x^2}=-2\left(\dfrac{\lim_{x\to0}\sin x}x\right)^2=?$$

Set $2x=2t,t$ and divide

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