The value of limx→∞ π2−tan−1⁡x1/x2, is

Let y=limx→∞ π2−tan−1⁡x1x 00 form ∴ log⁡y=limx→∞ 1xlog⁡π2−tan−1⁡x⇒ log⁡y=limx→∞ log⁡π/2−tan−1⁡xx                              ∞∞ form    ⇒ log⁡y=limx→∞ −11+x2π2−tan−1⁡x  [Using L'Hospital's Rule] ⇒log⁡y=limx→∞ 2x1+x22−11+x2 [ Using L'Hospital's Rule] ⇒ log⁡y=limx→∞ −2x1+x2=0⇒y=e0=1