Question

limit x tends to 2
1/x-2-2(2x-3)/x3-3x2+2x

Answer 1

Lim x→2 (x^2–5x+6)(x^2–3x+2)/(x^3–3x^2+4)=

=Limx→2 (x-6)(x+1)(x-1)(x-2)/[x(x-2)(x-1)+(-2x+4)]

x→2, x≠2, x=2+h where h→0

=Lim h→0 2(h-4)(h+3)(h+1)(h)/[(2+h)(h)(h+1)+(-2h)]

= Lim h→0 2(h)(h-4)(h+3)(h+1)/(h)[(2+h)(h+1)-2]

=Lim h→0 2(h-4)(h+3)(h-1)/[(2+h)(h+1)-2]

=Lim h→0 2(4)(3)(-1)/[2–2]=∞
Answer = ∞

Answer 2

lim x->2 $\frac{1}{x-2}$ -$\frac{2(2x-3)}{x^{3} -3x^{2}+2x }$ = -0.5

• (1/x-2)-(2(2x-3)/x^3-3x^2+2x)
• (1/x-2)-((2(2x-3)/x(x-1)(x-2))

Taking 1/x-2 common,

• 1/x-2*(1-(2(2x-3)/x(x-1))
• 1/x-2*(x^2-5x+6/x(x-1))
• 1/x-2*((x-2)(x-3)/x(x-1))

Cutting x-2, we get

• x-3/x(x-1)

Putting x=2 in the equation, the answer comes out to be -1/2 or -0.5