mathop {lim }limits_{x to 1} frac{{1 + cos pi | vedclass

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(B) Let $L = \mathop {\lim }\limits_{x 	o 1} \frac{{1 + \cos \pi x}}{{{{	an }^2}\pi x}}$.This is a $\frac{0}{0}$ form.Using $L$-Hospital's rule,we d

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$1/2$

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(B) Let $L = \mathop {\lim }\limits_{x 	o 1} \frac{{1 + \cos \pi x}}{{{{	an }^2}\pi x}}$.This is a $\frac{0}{0}$ form.Using $L$-Hospital's rule,we differentiate the numerator and denominator with respect to $x$:$L = \mathop {\lim }\limits_{x 	o 1} \frac{{-\pi \sin \pi x}}{{2 	an \pi x \cdot \sec^2 \pi x \cdot \pi}}$$L = \mathop {\lim }\limits_{x 	o 1} \frac{{-\sin \pi x}}{{2 	an \pi x \cdot \sec^2 \pi x}}$Since $	an \pi x = \frac{\sin \pi x}{\cos \pi x}$,we have:$L = \mathop {\lim }\limits_{x 	o 1} \frac{{-\sin \pi x \cdot \cos \pi x}}{{2 \sin \pi x \cdot \sec^2 \pi x}}$$L = \mathop {\lim }\limits_{x 	o 1} \frac{{-\cos \pi x}}{{2 \sec^2 \pi x}} = \mathop {\lim }\limits_{x 	o 1} \frac{{-\cos^3 \pi x}}{2}$Substituting $x = 1$:$L = \frac{{-\cos^3 \pi}}{2} = \frac{{-(-1)^3}}{2} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 1} \frac{{1 + \cos \pi x}}{{{{\tan }^2}\pi x}}$ is equal to

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