Math, asked by devaprbha2000, 6 months ago

Limit x tend to 3 (x^2-5x+6)/(2x^2-5x-3)

Answers

Answered by sahbajnumani
0

Answer:

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 =∞

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