Answer
$$\frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{\cos \left( {\pi /x} \right)}}{{x - 2}} \cr
& {\text{Using the hint }}\left[ {{\text{Let }}t = \frac{\pi }{2} - \frac{\pi }{x}} \right].{\text{ we have}} \cr
& x \to 2 \Rightarrow t = \frac{\pi }{2} - \frac{\pi }{2} = 0 \cr
& t \to 0{\text{ when }}x \to 2 \cr
& t = \frac{\pi }{2} - \frac{\pi }{x}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{\pi }{x} = \frac{\pi }{2} - t \cr
& and \cr
& x = \frac{{2\pi }}{{\pi - 2t}} \cr
& {\text{Substituting}} \cr
& \,\,\mathop {\lim }\limits_{x \to 2} \frac{{\cos \left( {\pi /x} \right)}}{{x - 2}} = \,\,\mathop {\lim }\limits_{t \to 0} \frac{{\cos \left( {\frac{\pi }{2} - t} \right)}}{{\frac{{2\pi }}{{\pi - 2t}} - 2}} \cr
& = \,\,\mathop {\lim }\limits_{t \to 0} \frac{{\cos \left( {\frac{\pi }{2} - t} \right)}}{{\frac{{4t}}{{\pi - 2t}}}} = \,\,\mathop {\lim }\limits_{t \to 0} \frac{{\left( {\pi - 2t} \right)\cos \left( {\frac{\pi }{2} - t} \right)}}{{4t}} \cr
& = \,\,\mathop {\lim }\limits_{t \to 0} \frac{{\left( {\frac{\pi }{2} - t} \right)\cos \left( {\frac{\pi }{2} - t} \right)}}{{2t}} \cr
& {\text{Use the Cofunction identity }}\cos \left( {\frac{\pi }{2} - x} \right) = \sin x.{\text{ we have}} \cr
& = \,\,\mathop {\lim }\limits_{t \to 0} \frac{{\left( {\frac{\pi }{2} - t} \right)\sin t}}{{2t}} \cr
& {\text{Use the product property of limits}} \cr
& = \,\,\left[ {\mathop {\lim }\limits_{t \to 0} \left( {\frac{\pi }{2} - t} \right)} \right]\left[ {\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{{2t}}} \right] \cr
& = \,\frac{1}{2}\,\left[ {\mathop {\lim }\limits_{t \to 0} \left( {\frac{\pi }{2} - t} \right)} \right]\left[ {\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t}} \right] \cr
& = \,\frac{1}{2}\,\left[ {\left( {\frac{\pi }{2} - 0} \right)} \right]\left[ 1 \right] \cr
& = \,\frac{\pi }{4} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{\cos \left( {\pi /x} \right)}}{{x - 2}} = \frac{\pi }{4} \cr} $$
