Tính giới hạn sau (mathop {lim }limits_{x to ,,3} dfrac{{sqrt {5x – 6} .sqrt[3]{{3x – 1}} – 2x}}{{{x^2} – x – 6}}).

(mathop {lim }limits_{x to ,,3} dfrac{{sqrt {5x – 6} .left( {sqrt[3]{{3x – 1}} – 2} right) + 2sqrt {5x – 6}  – 2x}}{{{x^2} – x – 6}}) ( = mathop {lim }limits_{x to ,,3} left[ {dfrac{{sqrt {5x – 6} .left( {sqrt[3]{{3x – 1}} – 2} right)}}{{{x^2} – x – 6}} + dfrac{{2sqrt {5x – 6}  – 2x}}{{{x^2} – x – 6}}} right])

( = mathop {lim }limits_{x to ,,3} left[ {dfrac{{3sqrt {5x – 6} left( {x – 3} right)}}{{left( {x – 3} right)left( {x + 2} right)left[ {{{left( {sqrt[3]{{3x – 1}}} right)}^2} + 2sqrt[3]{{3x – 1}} + 4} right]}} + dfrac{{2left( {x – 3} right)left( { – x + 2} right)}}{{left( {x – 3} right)left( {x + 2} right)left( {sqrt {5x – 6}  + x} right)}}} right])

( = mathop {lim }limits_{x to ,,3} left[ {dfrac{{3sqrt {5x – 6} }}{{left( {x + 2} right)left[ {{{left( {sqrt[3]{{3x – 1}}} right)}^2} + 2sqrt[3]{{3x – 1}} + 4} right]}} + dfrac{{2left( { – x + 2} right)}}{{left( {x + 2} right)left( {sqrt {5x – 6}  + x} right)}}} right])( = dfrac{1}{{12}})

Chọn C.