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

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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– cqfd
Jan 28, 2020 at 6:36

Only with limit laws and using

• $$\cos 2t = 2\cos^2 t -1$$

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

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 $$2\times1+2=4$$.

$$\cos (2t)=1-2\sin ^{2}(t)$$ 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 $$4$$.

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

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

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

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