Evaluate the Given limit: $$\lim\limits_{x \to 0} \dfrac{cos\;2x-1}{cos\;x-1}$$

Asked by Abhisek | 1 year ago |  112

##### Solution :-

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

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

Answered by Pragya Singh | 1 year ago

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