Question

lim x=0 (cos 2x-1)/(cos x -1)

Answer 1

4

Answer 2

The limit is of the form $\frac{0}{0}$. Using the trigonometric identities $\cos 2x = 1 - 2\sin^2 x$ and $\cos x = 1 - 2\sin^2(x/2)$, the expression simplifies to $\frac{-2\sin^2 x}{-2\sin^2(x/2)} = \frac{\sin^2 x}{\sin^2(x/2)}$. By multiplying and dividing by $x^2$ and $(x/2)^2$ respectively, and using the standard limit $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$, the limit evaluates to $\frac{1^2 \cdot x^2}{1^2 \cdot (x^2/4)} = 4$.