what would be LIMIT [1 / x-2 - 12 / x3+8] [X-TENDING TO (-2) ] - Maths - Limits and Derivatives

Question

what would be
LIMIT [1 / x-2 - 12 / x3
+8]
[X-TENDING TO (-2) ]

Anuradha Sharma · Accepted Answer

limx→-21x-2-12x3+8=limx→-2x3-12x+32x-2x3+8It is 00 form so using L'hospital rule we get, =limx→-2x3-12x+32x-2x3+8=limx→-2ddxx3-12x+32ddxx4+8x-2x3-16=limx→-23x2-124x3+8-6x2It is 00 form so using L'hospital rule we get,=limx→-26x12x2-12x=limx→-212x-2=-16

what would be LIMIT [1 / x-2 - 12 / x3 +8] [X-TENDING TO (-2) ]

what would be
LIMIT [1 / x-2 - 12 / x³ +8]
[X-TENDING TO (-2) ]