$\lim\limits_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}}$

Question

Exercise:

$$\lim\limits_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}}$$

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I've posted my solution down below, however if there are more elegant approaches, feel free to include your own solutions.

Answer 1

Step 1: find equivalent infinitesimals

Recall: $\lim\limits_{x \to 0}{\frac{\cos m - \cos n}{x^2}} = \frac{n^2 - m^2}{2}$. So: $\lim\limits_{x \to 0}{\frac{\cos x - \cos 2x}{x^2}} = \frac{2^2 - 1^2}{2} = \frac{3}{2}$. To make $RHS = 1$: $\lim\limits_{x \to 0}{\frac{\cos x - \cos 2x}{\frac{3}{2}x^2}} = 1$. Finally: $\cos x - \cos 2x \sim \frac{3}{2}x^2$.

Recall: $\lim\limits_{x \to 0}{\frac{1 - \cos x}{x^2}} = \frac{1}{2}$. To make $RHS = 1$: $\lim\limits_{x \to 0}{\frac{1 - \cos x}{\frac{1}{2}x^2}} = 1$. Finally: $1 - \cos x \sim \frac{1}{2}x^2$

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Step 2: replace with equivalent infinitesimals

$\lim\limits_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}}$ $= \lim\limits_{x \to 0}{\frac{\frac{3}{2}x^2}{\frac{1}{2}x^2}}$ $= 3$

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Answer:

$$\lim\limits_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}} = 3$$

Answer 2

By applying L'Hospital's rule twice, one has $$ \lim\limits_{x \to 0}{\frac{\cos x - \cos 2x}{1 - \cos x}}=\lim\limits_{x \to 0}{\frac{-\sin x +2 \sin 2x}{\sin x}}=\lim\limits_{x \to 0}{\frac{-\cos x +4 \cos 2x}{\cos x}}=\frac{-1+4}1=3. $$

Answer 3

We have the double angle formula:

$$\cos(2x)=2\cos^2(x)-1$$

Thus, it follows that

$$\cos(x)-\cos(2x)=1+\cos(x)-2\cos^2(x)=(1-\cos(x))(2\cos(x)+1)$$

Thus, we find

$$\frac{\cos(x)-\cos(2x)}{1-\cos(x)}=2\cos(x)+1$$

So the limit is directly $3$.

Answer 4

$=\dfrac{\cos x - \cos 2x}{1-\cos x}$

$=\dfrac{2\sin \left( \frac{x+2x}{2}\right) \sin \left( \frac{2x-x}{2}\right) }{2\sin^2 \left( \frac{x}{2}\right)}$

$=\dfrac{\sin 3x/2}{x} . \dfrac{x}{\sin x/2}$

$=\left(3/2\right) . \frac{1}{\left( 1/2\right)}$

$=3$

Formulae involved--

$\cos A - \cos B = 2\sin \frac{A+B}{2} \sin \frac{B-A}{2}$

$\cos 2A = 1-2\sin^2 A$

$\lim\limits_{x\to 0} \dfrac {\sin ax}{x} =a$

Answer 5

If we evaluate directly we get the fraction $\frac{0}{0}$. So using L'Hopital's rule: $$ \lim_{x \rightarrow 0} \frac{\cos x - \cos 2x}{1-\cos x} = \lim_{x \rightarrow 0} \frac{-\sin x + 2\sin 2x}{\sin x} $$

Now, if we evaluate directly once more, we get the fraction $\frac{0}{0}$. So we use L'Hopital again to get:

$$ \lim_{x \rightarrow 0} \frac{-\cos x + 4 \cos 2x}{\cos x} $$

Which, when evaluated on $x=0$ yields: $$\frac{-1 + 4}{1} = 3$$