Evaluate lim x → ∞ ( 1 x + 2 1 x ) x

Desirae Washington

Desirae Washington

Answered

2022-07-13

Evaluate lim x ( 1 x + 2 1 x ) x

Answer & Explanation

Feriheashrz

Feriheashrz

Expert

2022-07-14Added 8 answers

You have that
e 1 + x 2 1 x x = e 1 + x ( 2 1 x 1 )
Now recall that
lim t 0 a t 1 t = lim t 0 e t log a 1 t = lim t 0 e t log a 1 t log a log a
By putting u = t log a and noticing that t log a 0 as t 0, hence
lim t 0 e t log a 1 t log a log a = lim u 0 e u 1 u log a = 1 log a = log a
So, in e 1 + x ( 2 1 x 1 ) , putting z = 1 x you have that z 0 as x and so
lim x e 1 + x ( 2 1 x 1 ) = lim z 0 + e 1 + 2 z 1 z = e 1 + log 2 = e e log 2 = 2 e
Alternatively, notice that since when x it is 1 x + 2 1 / x > 0, it is
( 1 x + 2 1 / x ) x = exp [ x log ( 1 x + 2 1 / x ) ] = exp { x [ 1 x log 2 + log ( 1 + 1 x 2 1 / x ) ] }
Now use that log ( 1 + t ) = t + o ( t ) as t 0; I'm using exp [ f ( x ) ] = e f ( x ) for better formatting.
Shea Stuart

Shea Stuart

Expert

2022-07-15Added 4 answers

Taking the logarithm of the function and setting t := 1 x , we can use L'Hospital:
lim t 0 log ( t + 2 t ) t = lim t 0 ( 1 + 2 t log 2 ) t + 2 t .
= 1 + log 2, giving the final answer 2 e

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