If cos−1⁡1n<π2, then limn→∞ (n+1)2πcos−1⁡1n−n is equal to

limn→∞ (n+1)2πcos−1⁡1n−n=limn→∞ 2π(n+1)cos−1⁡1n−π2n=limn→∞ 2πncos−1⁡1n−π2+cos−1⁡1n=limn→∞ 2π−nsin−1⁡1n+cos−1⁡1n=limn→∞ 2π−sin−11n1n+cos−11n=2π−1+π2=π−2π