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7.2 Q.III · Q33

Q.lim⁡x→1[x−2x2−x−1x3−3x2+2x]\displaystyle\lim_{x\to 1}\left[\frac{x-2}{x^2-x}-\frac{1}{x^3-3x^2+2x}\right]

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x2−x=x(x−1)x^2-x=x(x-1) and x3−3x2+2x=x(x−1)(x−2)x^3-3x^2+2x=x(x-1)(x-2). Writing the first term as (x−2)2x(x−1)(x−2)\dfrac{(x-2)^2}{x(x-1)(x-2)} and combining with −1x(x−1)(x−2)-\dfrac{1}{x(x-1)(x-2)}: numerator =(x−2)2−1=(x−3)(x−1)=(x-2)^2-1=(x-3)(x-1). So the expression is (x−3)(x−1)x(x−1)(x−2)=x−3x(x−2)\dfrac{(x-3)(x-1)}{x(x-1)(x-2)}=\dfrac{x-3}{x(x-2)} after cancelling (x−1)(x-1). At x=1x=1: −21×(−1)=2\dfrac{-2}{1\times(-1)}=2.

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