I have one question about limits: it is required to find the limit <mstyle displaystyle="true" sc

grenivkah3z 2022-07-10 Answered
I have one question about limits: it is required to find the limit lim n 3 n + 2 n n
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Answers (2)

Isla Klein
Answered 2022-07-11 Author has 12 answers
Notice that
( 3 n + 2 n ) 1 / n = ( 5 n 3 n + 2 n 5 n ) 1 / n = 5 [ ( 3 5 ) n + ( 2 5 ) n ] 1 / n .
Thus, when you concluded that the second limit is 0 you virtually assumed that
lim n [ ( 3 5 ) n + ( 2 5 ) n ] 1 / n = 0 ,
which is false. Your assumption is that since
lim n ( 3 5 ) n + ( 2 5 ) n = 0 ,
that the former limit must also be 0 However, this is not the case: in the cases where the exponent has limit 0 it is not sufficient for the base to have limit 0 in order for the entire power to have limit 0 The exponent is 1 / n , and notice that
lim n 1 n = 0.
As such, your second calculation is simply incorrect. In general, if
lim n a n = 0
and
lim n b n = 0 ,
then
lim n b n a n
can be equal to any nonnegative real number, or it can even not exist. You cannot simply conclude it to be 0.
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babyagelesszj
Answered 2022-07-12 Author has 7 answers
The expression just equals:-
lim n 3 ( 1 + ( 2 3 ) n ) 1 n
Now this is 1 form.
You can use this whenever you run into this 1 form
lim x a f ( x ) g ( x ) = exp ( lim x a ( f ( x ) 1 ) g ( x ) )
when lim x a f ( x ) g ( x ) is 1 form.
So using this we get it as :-
3 exp ( lim n ( 2 3 ) n n ) = 3 exp ( 0 ) = 3
In general whenever you have something like
lim n ( a 1 n + a 2 n + . . . + a m n ) 1 n
where a i 's are distinct positive real numbers.You can use the above method and immediately write the answer as max 1 i m a i
That is we are strongly using the fact that a n 0 when 0 < a < 1
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