We have,
limx→1((2x−3)(√x−1)3x2+3x−6)
This is 00 form.
So, apply L-Hospital rule,
limx→1⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝(2−0)(√x−1)+(2x−3)(12√x−0)6x+3−0⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠
limx→1⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝2(√x−1)+(2x−3)(12√x)6x+3⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠
=2(1−1)+(2−3)(12)6+3
=−129
=−118
Hence, the value is −118.