n → ∞ lim n x − ( n − 1 ) x n 1 9 9 9 = 2 0 0 0 1 , x = ?
Clarification: Take x as a constant.
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Since this is currently of the form (infinity/infinity), we may use LH rule of differentiation:
Limit x ( n x − 1 − ( n − 1 ) x − 1 ) 1 9 9 9 n 1 9 9 8
Let us assume x>1999. So we keep differentiating by LH rule till power of n in numerator become 0, so we get:
Limit x ( x − 1 ) ( x − 2 ) . . . ( x − 1 9 9 8 ) ( n x − 1 9 9 9 − ( n − 1 ) x − 1 9 9 9 ) 1 9 9 9 ∗ 1 9 9 8 ∗ 1 9 9 7 . . . . 2 ∗ 1
Now since we still have a term of n in denominator we will get the value of limit =0 as n become infinite. But value of limit is 1/2000, so the power of n in denominator should also vanish, for which x=2000.
We may also check, in case x=2000, limit becomes
2 0 0 0 ∗ 1 9 9 9 ∗ 1 9 9 8 . . . 3 ∗ 2 ( n − ( n − 1 ) 1 9 9 9 ∗ 1 9 9 8 ∗ 1 9 9 7 . . . . 2 ∗ 1
=
1999!/2000! = 1/2000
The same method can be used by assuming x<1999 and proceeding. In that case power of n will vanish first in denominator and then will have have to make power of n = 0 in numerator so that limit becomes finite. It will again yield x=2000
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n → ∞ lim n x − ( n − 1 ) x n 1 9 9 9 ⇒ x = 2 0 0 0 = n → ∞ lim n x − ( n x − x n x − 1 + 2 x ( x − 1 ) n x − 2 − . . . ) n 1 9 9 9 = n → ∞ lim x n x − 1 − 2 x ( x − 1 ) n x − 2 + 6 x ( x − 1 ) ( x − 2 ) n x − 3 − . . . n 1 9 9 9 = n → ∞ lim x n x − 2 0 0 0 − 2 x ( x − 1 ) n x − 2 0 0 1 + 6 x ( x − 1 ) ( x − 2 ) n x − 2 0 0 2 . . . 1 When x = 2 0 0 0 = n → ∞ lim 2 0 0 0 − 2 n 2 0 0 0 ( 1 9 9 9 ) + 6 n 2 2 0 0 0 ( 1 9 9 9 ) ( 1 9 9 8 ) . . . 1 = 2 0 0 0 1