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2d3vljtq 2022-07-10 Answered
Evalutating lim x ( x + 1 ) k ( x ) k , given 0 < k < 1
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Answers (2)

haingear8v
Answered 2022-07-11 Author has 13 answers
You can just rewrite the expression and use L'Hôpital's rule:
lim x + ( ( x + 1 ) k x k ) = lim x + ( 1 + 1 / x ) k 1 1 / x k = lim x + ( k / x 2 ) ( 1 + 1 / x ) k 1 k x k 1 = lim x + ( x + 1 ) k 1 = 0
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civilnogwu
Answered 2022-07-12 Author has 6 answers
Use Taylor polynomial: as h 0, we have ( 1 + h ) k = 1 + k h + o ( h )
As x
( x + 1 ) k = x k ( 1 + 1 x ) k = x k ( 1 + k x + o ( 1 / x ) ) = x k + k x k 1 + o ( x k 1 ) ( x + 1 ) k x k = k x k 1 + o ( x k 1 )
Now k < 1 so k 1 < 0 and thus lim x x k 1 = 0. Conclude
( x + 1 ) k x k 0  as  x
or more pricisely
( x + 1 ) k x k k x k 1  as  x
Example:
x + 1 x 1 2 x  as  x
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