Question

limit x tends to 2/3 [3/2-3x + 2/x²-2x] evaluate the limit​

Answer 1

Answer:

Lim(x→2) [ 1/(x - 2) - 2(2x-3)/(x³ - 3x² + 2x ) ]

Lim(x→2) [ 1/(x - 2) - 2(2x - 3)/x(x² - 3x + 2)]

Lim(x→2) [1/(x - 2) - 2(2x - 3)/x(x-2)(x -1) ]

Lim(x→2) [{x(x - 1) - 2(2x - 3)}/x(x - 1)(x - 2)]

Lim(x→2) [{x² - x - 4x + 6 }/x(x -1)(x -2)]

Lim(x→2) (x² - 5x + 6)/x(x - 1)(x -2)

Lim(x→2) (x -2)(x -3)/x(x -1)(x -2)

Lim(x→2) (x -3)/x(x -1)

now put x = 2 in Limit

= (2 - 3)/2(2-1) = -1/2