Evaluate the Given limit:limits_{x \to 0}{cos2x-1}{cosx-1}

\( \lim\limits_{x \to 0} \dfrac{cos\;2x-1}{cos\;x-1} \)

At x = 0, the value of the given function takes the form \( \dfrac{0}{0}\)

= \( \lim\limits_{x \to 0} \dfrac{1-2sin^2x-1}{1-2sin^2\dfrac{x}{2}-1} \)

= \( \lim\limits_{x \to 0} \dfrac{\dfrac{sin^2x}{x^2}\times x^2}{\dfrac{sin^2\dfrac{x}{2}}{(\dfrac{x}{2})^2}\times \dfrac{x^2}{4}} \)

= \( 4 \lim\limits_{x \to 0} \dfrac{\dfrac{sin^2x}{x^2}} { \lim\limits_{x \to 0}\dfrac{sin^2\dfrac{x}{2}}{(\dfrac{x}{2})^2}} \)

= \( 4 \lim\limits_{x \to 0} \dfrac{\dfrac{sin^2x}{x^2}} { \lim\limits_{x \to 0}\dfrac{sin^2\dfrac{x}{2}}{(\dfrac{x}{2})^2}}\)

= \(4 \dfrac{1^2}{1^2}\)   \( [\lim\limits_{y \to 0} (\dfrac{sin\;y}{y})=1]\)

= 4