Find the limit $\lim_{x\to 0}\frac{1-\cos 2x}{1-\cos x}$ without using L'Hospital

Question

Help me calculate this limit:

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

Answer 1

Hint: $$\frac{1-\cos 2x}{1-\cos x} = \frac{2\sin^2 x}{2\sin^2\frac{x}2}.$$

Answer 2

1. Use L'Hospital twice $$ \lim_{x \to 0} \frac {1-\cos 2x}{1 - \cos x} = \lim_{x \to 0} \frac {\left ( 1-\cos 2x \right )'}{\left( 1 - \cos x \right )'} = \lim_{x \to 0} \frac {2\sin 2x}{\sin x} = \lim_{x \to 0} \frac {\left ( 2\sin 2x \right )'}{\left ( \sin x \right )'} = \lim_{x \to 0} \frac {4\cos 2x}{\cos x} = 4 $$

2. Use Taylor series expansion of $\cos x$ around $0$ $$ \lim_{x \to 0} \frac {1-\cos 2x}{1-\cos x} = \lim_{x \to 0} \frac {1-1+\frac {4x^2}2 + O((2x)^4)}{1-1+\frac {x^2}2 + O(x^4)} = 4 $$

3. Use double angle formula $$ \lim_{x \to 0} \frac {1-\cos 2x}{1-\cos x} = \lim_{x \to 0} \frac {2 - 2 \cos^2x}{1-\cos x} = \lim_{x \to 0} \frac {2(1-\cos x)(1+\cos x)}{1-\cos x} = \lim_{x \to 0} 2(1+\cos x) = 4 $$

Answer 3

Use $\cos 2x= \cos^2 x-\sin^2x$, which gives $1-\cos 2x=2\sin^2 x = 2(1-\cos^2x)$ and you can cancel the denominator.