We have,
limx→0(sin(6x2)ln(cos(2x2−x)))
This is the 00 form.
So, apply L-Hospital rule
limx→0⎛⎜ ⎜ ⎜ ⎜⎝cos(6x2)×(12x)1cos(2x2−x)×(−sin(2x2−x))×(4x−1)⎞⎟ ⎟ ⎟ ⎟⎠
limx→0(12xcos(6x2)−tan(2x2−x)×(4x−1))
This is the 00 form.
So, apply L-Hospital rule
limx→0(12xcos(6x2)−(4x−1)tan(2x2−x))
limx→0(12cos(6x2)+12x(−sin(6x2))×12x−(4x−1)sec2(2x2−x)×(4x−1)−tan(2x2−x)(4−0))
limx→0(12cos(6x2)−144x2sin(6x2)−(4x−1)2sec2(2x2−x)−4tan(2x2−x))
=12cos0−0−(0−1)2sec20−4tan0
=12−1−0
=−12
Hence, this is the answer.