Finding limits at infinity?

Question

what is the limit as x approaches positive infinity for the function (xe^(1/x)-x)(e^x+x)^(1/x)

Anyone have any idea how to do this? I'd appreciate any help.

Thanks.

Answer 1

as x approaches infinity then 1/x = zero, and e^1/x = 1

As x approaches infinity then, xe^1/x -x = x [ e^1/x -1 ] = zero

Also, as x approaches infinity then(e^x + x) ^ !/x = 1, because any finite number raised to the power zero is equal to one.

Therefore the final answer is zero.

Answer 2

the first factor can be written as
[e^(1/x)-1)/1/x which is known to have lim 1
Write it as (e^z-1)/z for z==>0= lim e^z/1 =1
so remains lim (e^x+x)^1/x
take ln you get
1/x * ln[e^x)(1+x/e^x)= 1+1/xln(1+x/e^x) ===> 1 as x/e^x==>0
if the ln ==>1 the function ==> e and the limit is e