If f(x)=2x−3,g(x)=x−3x+4 and h(x)=−2(2x+1)x2+x−12 then lim 7x→3 {f(x)+g(x)+h(x)} is

If f(x)=2x3,g(x)=x3x+4 and h(x)=2(2x+1)x2+x12 then lim 7x3{f(x)+g(x)+h(x)} is

  1. A
  2. B
  3. C
  4. D

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    Solution:

    We have,

    f(x)+g(x)+h(x)=x24x+174x2x2+x12=(x3)(x5)(x3)(x+4) limx3[f(x)+g(x)+h(x)]=limx3(x3)(x5)(x3)(x+4)=27

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