limx→π/6 3sin⁡x−3cos⁡x6x−π equals

We have,limx→π/6 3sin⁡x−3cos⁡x6x−π=limx→π/6 2332sin⁡x−12cos⁡x6x−π6=limx→π/6 13sin⁡x−π6x−π6=13ALITER We have,limx→π/6 3sin⁡x−3cos⁡x6x−π 00 form =limx→π/6 3cos⁡x+3sin⁡x6               [Using De L Hospital's Rule] =163cos⁡π6+3sin⁡π6=13