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

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Question

(B) ધારો કે $L = \mathop {\lim }\limits_{x 	o 1} \frac{{1 + \cos \pi x}}{{{{	an }^2}\pi x}}$.આ $\frac{0}{0}$ સ્વરૂપ છે.$L$-Hospital ના નિયમનો ઉપયોગ

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(B) ધારો કે $L = \mathop {\lim }\limits_{x 	o 1} \frac{{1 + \cos \pi x}}{{{{	an }^2}\pi x}}$.આ $\frac{0}{0}$ સ્વરૂપ છે.$L$-Hospital ના નિયમનો ઉપયોગ કરતા:$L = \mathop {\lim }\limits_{x 	o 1} \frac{{-\pi \sin \pi x}}{{2 	an \pi x \cdot \sec^2 \pi x \cdot \pi}}$$L = \mathop {\lim }\limits_{x 	o 1} \frac{{-\sin \pi x}}{{2 	an \pi x \cdot \sec^2 \pi x}}$$	an \pi x = \frac{\sin \pi x}{\cos \pi x}$ હોવાથી:$L = \mathop {\lim }\limits_{x 	o 1} \frac{{-\cos^3 \pi x}}{2}$$x = 1$ મુકતા:$L = \frac{{-\cos^3 \pi}}{2} = \frac{{-(-1)^3}}{2} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 1} \frac{{1 + \cos \pi x}}{{{{\tan }^2}\pi x}}$ ની કિંમત શોધો.

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