Question

Solve
${lim}_{x\to 0}\left(\frac{cos2x-1}{cosx-1}\right)$.

Answer 1

$\underset{x\to 0}{lim}\frac{\cos 2x-1}{\cos x-1}=$indeterminate form

Applying L'Hospital Rule

$\underset{x\to 0}{lim}\frac{-2\sin 2x}{-sinx}=$indeterminate form

Applying L'Hospital rule

$\underset{x\to 0}{lim}\frac{4\cos 2x}{\cos x}=\frac{4}{1}=4$