Evaluate the given limit: mathop {lim }limits | vedclass

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(B) We need to evaluate $\mathop {\lim }\limits_{x 	o 0} \frac{\cos 2x - 1}{\cos x - 1}$.At $x = 0$,the expression takes the indeterminate form $\fra

Vedclass · Accepted Answer

$4$

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(B) We need to evaluate $\mathop {\lim }\limits_{x 	o 0} \frac{\cos 2x - 1}{\cos x - 1}$.At $x = 0$,the expression takes the indeterminate form $\frac{0}{0}$.Using the trigonometric identity $\cos 2	heta = 1 - 2\sin^2 	heta$,we have:$\cos 2x = 1 - 2\sin^2 x$ and $\cos x = 1 - 2\sin^2 \frac{x}{2}$.Substituting these into the limit:$\mathop {\lim }\limits_{x 	o 0} \frac{(1 - 2\sin^2 x) - 1}{(1 - 2\sin^2 \frac{x}{2}) - 1} = \mathop {\lim }\limits_{x 	o 0} \frac{-2\sin^2 x}{-2\sin^2 \frac{x}{2}} = \mathop {\lim }\limits_{x 	o 0} \frac{\sin^2 x}{\sin^2 \frac{x}{2}}$.Dividing numerator and denominator by $x^2$ and multiplying by appropriate factors:$= \mathop {\lim }\limits_{x 	o 0} \frac{(\frac{\sin x}{x})^2}{(\frac{\sin(x/2)}{x/2})^2 \cdot \frac{1}{4}} = \frac{1^2}{1^2 \cdot \frac{1}{4}} = 4$.

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}$

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