I have the following limit question, where different indices of roots appear in the numerator and the denominator
limx→33√x+5−24√x−2−1.
As we not allowed to use L'Hopital, I want to learn how we can proceed algebraically.
Answer
A possible purely algebraic way can use
an−bn=(a−b)(an−1+an−2b+⋯+abn−2+bn−1)
Using this formula note that
- 4√x−2−1=(4√x−2)4−14(4√x−2)3+(4√x−2)2+4√x−2+1=x−3(4√x−2)3+(4√x−2)2+4√x−2+1
- 3√x+5−2=(3√x+5)3−23(3√x+5)2+3√x+5⋅2+22=x−3(3√x+5)2+3√x+5⋅2+22
So, you get
3√x+5−24√x−2−1=(4√x−2)3+(4√x−2)2+4√x−2+1(3√x+5)2+3√x+5⋅2+22x→3⟶412=13
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