Evaluate the following limits:

Question:

Evaluate the following limits:

$\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}$

 

Solution:

$=\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}$

$=\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1} \times \frac{4 \times x \times x}{2 x \times 2 x}$

$=4 \times \lim _{x \rightarrow 0} \frac{\frac{\cos 2 x-1}{2 x \times 2 x}}{\frac{\cos x-1}{x \times x}}$

$=4 \times \frac{\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{(2 x)^{2}}}{\lim _{x \rightarrow 0} \frac{\cos x-1}{x^{2}}}$

$=4 \times \frac{\frac{1}{2}}{\frac{2}{2}}=4$

$\therefore \lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}=4$

 

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