The value of  lim_(xrarr2) (cos (π/x))/(x-2)   is -- 

প্রদত্ত, \( \lim_{x \to 2} \frac{\cos(\frac{\pi}{x})}{x-2} \)

ধরি, \( x = 2 + h \) সুতরাং, যখন \( x \to 2 \), \( h \to 0 \)

সুতরাং, \( \lim_{x \to 2} \frac{\cos(\frac{\pi}{x})}{x-2} = \lim_{h \to 0} \frac{\cos(\frac{\pi}{2+h})}{h} \)

আমরা জানি, \( \cos(A) = -\sin(A - \frac{\pi}{2}) \)

সুতরাং, \( \cos(\frac{\pi}{2+h}) = - \sin(\frac{\pi}{2+h} - \frac{\pi}{2}) = - \sin(\frac{2\pi - \pi(2+h)}{2(2+h)}) = - \sin(\frac{-\pi h}{2(2+h)}) = \sin(\frac{\pi h}{2(2+h)}) \)

তাহলে, \( \lim_{h \to 0} \frac{\cos(\frac{\pi}{2+h})}{h} = \lim_{h \to 0} \frac{\sin(\frac{\pi h}{2(2+h)})}{h} \)

আমরা জানি, \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

সুতরাং, \( \lim_{h \to 0} \frac{\sin(\frac{\pi h}{2(2+h)})}{h} = \lim_{h \to 0} \frac{\sin(\frac{\pi h}{2(2+h)})}{\frac{\pi h}{2(2+h)}} \cdot \frac{\pi}{2(2+h)} = 1 \cdot \lim_{h \to 0} \frac{\pi}{2(2+h)} \)

\( = \frac{\pi}{2(2+0)} = \frac{\pi}{4} \)

অতএব, \( \lim_{x \to 2} \frac{\cos(\frac{\pi}{x})}{x-2} = \frac{\pi}{4} \) 🥳