Evaluate the following limits in limx→π2tan 2xx−π2
Here limx→π2tan 2xx−π2[00form]
Put x = π2+ y as x→π2,y→0
∴limy→0tan 2(π2+y)π2+y−π2limy→0tan(π+2y)y
= limy→0tan2yy
= limy→0tan2y2y×2
= 1 × 2 = 2.
Evaluate: limx→π2tan2xx−π2