displaystyle limx → 2[(1/x-2)-(2(2x-3)/x3-3x2+2x)] is equal to

Question

displaystyle limx → 2[(1/x-2)-(2(2x-3)/x3-3x2+2x)] is equal to (A) -(1/2) (B) (1/2) (C) 0 (D) ∞

Tardigrade Admin · Accepted Answer

We have, displaystyle limx → 2[(1/x-2)-(2(2x-3)/x3-3x2+2x)] = displaystyle limx → 2[(1/x-2)-(2(2x-3)/x(x-1)(x-2))] = displaystyle limx → 2[(x(x-1)-2(2x-3)/x(x-1)(x-2))] = displaystyle limx → 2[(x2-5x+6/x(x-1)(x-2))] = displaystyle limx → 2[((x-2)(x-3)/x(x-1)(x-2))] = displaystyle limx → 2[((x-3)/x(x-1))]=(-1/2)

Q. $x \to 2 lim$ $[\frac{1}{x - 2} - \frac{2 ( 2 x - 3 )}{x ^{3} - 3 x ^{2} + 2 x}]$ is equal to

$- \frac{1}{2}$

$\frac{1}{2}$

$0$

$\infty$

Q. x→2lim​[x−21​−x3−3x2+2x2(2x−3)​] is equal to

−21​

21​

0

∞

We have, x→2lim​[x−21​−x3−3x2+2x2(2x−3)​] =x→2lim​[x−21​−x(x−1)(x−2)2(2x−3)​] =x→2lim​[x(x−1)(x−2)x(x−1)−2(2x−3)​] =x→2lim​[x(x−1)(x−2)x2−5x+6​] =x→2lim​[x(x−1)(x−2)(x−2)(x−3)​] =x→2lim​[x(x−1)(x−3)​]=2−1​

Q. $x \to 2 lim$ $[\frac{1}{x - 2} - \frac{2 ( 2 x - 3 )}{x ^{3} - 3 x ^{2} + 2 x}]$ is equal to

$- \frac{1}{2}$

$\frac{1}{2}$

$0$

$\infty$