The value of mathop {lim }limits_{x to 1} fra | vedclass

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(C) Let the expression be $L = \mathop {\lim }\limits_{x 	o 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$.Using t

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$\frac{2}{\sin 2}$

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(C) Let the expression be $L = \mathop {\lim }\limits_{x 	o 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$.Using the identity $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$,we have $\cos x \cos (x+2) = \frac{1}{2} [\cos(2x+2) + \cos(-2)] = \frac{1}{2} [\cos(2x+2) + \cos 2]$.Also,${{\sin }^2}x - {{\cos }^2}(x+1) = \frac{1 - \cos 2x}{2} - \frac{1 + \cos(2x+2)}{2} = \frac{-\cos 2x - \cos(2x+2)}{2}$.Substituting these,the denominator becomes $\frac{1}{2} [-\cos 2x - \cos(2x+2) + \cos(2x+2) + \cos 2] = \frac{\cos 2 - \cos 2x}{2} = \frac{2 \sin(\frac{2x+2}{2}) \sin(\frac{2x-2}{2})}{2} = \sin(x+1) \sin(x-1)$.Thus,$L = \mathop {\lim }\limits_{x 	o 1} \frac{(x-1)(x+1)}{\sin(x+1) \sin(x-1)} = \mathop {\lim }\limits_{x 	o 1} \left( \frac{x-1}{\sin(x-1)} ight) \cdot \frac{x+1}{\sin(x+1)} = 1 \cdot \frac{2}{\sin 2} = \frac{2}{\sin 2}$.

The value of $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$ is:

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