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7.2 Q.II · Q28

Q.lim⁡x→2[x3−4x2+4xx2−1]\displaystyle\lim_{x\to 2}\left[\frac{x^3-4x^2+4x}{x^2-1}\right]

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Numerator x3−4x2+4x=x(x−2)2x^3-4x^2+4x=x(x-2)^2; at x=2x=2 this is 2(0)2=02(0)^2=0. Denominator x2−1x^2-1 at x=2x=2 is 33, non-zero. So no cancellation is even needed …

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