Question

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

• A. $-\frac{1}{4}$
• B. $\frac{1}{5}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{1}{4}$

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

$=\underset{x\to 0}{lim}\frac{-2{\sin }^{2}x}{-2{\sin }^2\left(\frac{x}{2}\right)}$

$=\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{x^2}\frac{x^2}{{\sin }^2\left(\frac{x}{2}\right)}$

$=\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^2\frac{1}{4\left(\frac{{\sin }^2\left(\frac{x}{2}\right)}{\frac{x^2}{4}}\right)}$

$=\underset{x\to 0}{lim}\frac{1}{4}{\left(\frac{\sin x}{x}\right)}^2\frac{1}{{\left(\frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)}^2}$

$=\frac{1}{4}\left(1\right)\left(1\right)$

$=\frac{1}{4}$