Question:
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}$
Solution:
$=\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}$
$=\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1} \times \frac{4 \times x \times x}{2 x \times 2 x}$
$=4 \times \lim _{x \rightarrow 0} \frac{\frac{\cos 2 x-1}{2 x \times 2 x}}{\frac{\cos x-1}{x \times x}}$
$=4 \times \frac{\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{(2 x)^{2}}}{\lim _{x \rightarrow 0} \frac{\cos x-1}{x^{2}}}$
$=4 \times \frac{\frac{1}{2}}{\frac{2}{2}}=4$
$\therefore \lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}=4$