limx→∞ x232−232x+416−1(x−1)2 is equal to

limxx232232x+4161(x1)2 is equal to

  1. A

    263-231

  2. B

    264-231

  3. C

    262-231

  4. D

    265-233

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    Solution:

       limx1x23232x+4161(x1)2

    =limx1xnnx+n1(x1)2, where n=232

    =limx1xn1n(x1)(x1)2=limx1xn1+xn2+xn3++x+1nx1

    =limx1xn11+xn21+xn31++(x1)x1=limx1xn11x1+xn21x1+xn31x1++x1x1={(n1)+(n2)+(n3)++2+1}

    =n(n1)2=2312321=263231

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