So i have two limits that i want to solve, so i will put them together in one question, while both are probably solved with help of L'Hospital's rule. I am not sure for the second one.

Julius Blankenship 2022-09-14 Answered
So i have two limits that i want to solve, so i will put them together in one question, while both are probably solved with help of L'Hospital's rule. I am not sure for the second one. 1)
lim x 0 ( e 4 ( 1 x 2 ) 1 x 2 ( 1 2 x ) 1 x ) 1 x
I am having problems with this one to make it into one that can be used for L'Hospital's rule, but i cannot get it into 0 0 , there are others options as well, but i am stuck.
2) lim n ( arcsin ( 1 n ) + arcsin ( 2 n ) + . . . + arcsin ( n n ) n )
So i thought i could solve this one even if i used comparing test, but i cannot fins a way, but this one is probably not solved with L'Hospital rule, because of n, but i put it in one question, because i found both limits in the same group. I apologize for putting them in the same question if they differ too much. Any help would be appreciated.
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Answers (2)

Isaiah Haynes
Answered 2022-09-15 Author has 16 answers
Step 1
For the first if x 0 then the limit does not exist. To prove this let us observe that
( 1 x 2 ) 1 / x 2 e 1 , ( 1 2 x ) 1 / x e 2
as x 0 and hence
f ( x ) = ( e 4 ( 1 x 2 ) 1 / x 2 ( 1 2 x ) 1 / x ) e 5 > 1
and g ( x ) = 1 / x as x 0 + so { f ( x ) } g ( x ) as x 0 + .
Step 2
But when x 0 then g ( x ) and hence { f ( x ) } g ( x ) 0 as x 0 . It follows that the desired limit does not exist. In the above we have used the following standard limits
lim x 0 ( 1 + a x ) 1 / x = e a , a x  as  x  if  a > 1
The second limit is a Riemann sum and can be evaluated easily (as explained in answer from alans).

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yamyekay3
Answered 2022-09-16 Author has 1 answers
Step 1
As regards the first limit, by using
lim t 0 ( 1 + a t ) 1 / t = lim s + ( 1 + a / s ) s = e a we have that
lim x 0 ( e 4 ( 1 x 2 ) 1 x 2 ( 1 2 x ) 1 x ) = e 4 e 1 e 2 = e 5 .
So the base of the exponential is eventually greater then 2 (something > 1 and less than e 5 ).
Step 2
If x 0 + then for 0 < x < r for some r > 0,
+ 2 1 / x < ( e 4 ( 1 x 2 ) 1 x 2 ( 1 2 x ) 1 x ) 1 x
If x 0 then for r < x < 0 for some r > 0,
0 < ( e 4 ( 1 x 2 ) 1 x 2 ( 1 2 x ) 1 x ) 1 x < 2 1 / x 0.
Therefore you have different behaviours for x 0 and x 0 + and the limit for x 0 does not exist!

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