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7.2 Q.III · Q34

Q.lim⁡x→1[x4−3x2+2x3−5x2+3x+1]\displaystyle\lim_{x\to 1}\left[\frac{x^4-3x^2+2}{x^3-5x^2+3x+1}\right]

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x4−3x2+2=(x2−1)(x2−2)=(x−1)(x+1)(x2−2)x^4-3x^2+2=(x^2-1)(x^2-2)=(x-1)(x+1)(x^2-2). For the denominator, synthetic division of x3−5x2+3x+1x^3-5x^2+3x+1 by (x−1)(x-1) gives quotient x2−4x−1x^2-4x-1, so x3−5x2+3x+1=(x−1)(x2−4x−1)x^3-5x^2+3x+1=(x-1)(x^2-4x-1). Cancelling (x−1)(x-1): $\dfrac{(x+1)(x^2-2)}{x^2 …

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