Question

Find \[\lim_{x \rightarrow 0}\frac{ e ^{\frac{ 1 }{ x }}-1 }{ e ^{\frac{ 1 }{ x }}+1 }\]

Answer 1

Dont Use L-Hospitals Rule

Answer 2

try supposed t = 1/x
if x=0 -> t=~
so, it can be :
lim (t->~) (e^t - 1)/(e^t + 1)
= lim (t->~) (e^t + 1 - 2)/(e^t + 1)
= lim (t->~) 1 - 2/(e^t + 1)

continue it ...

Answer 3

Ummm. Isnt this straight away 1? I'm missing something I think.

Answer 4

no, it already make sense.. just calculate the limit of 2/(e^t + 1) for t -> ~
and obviously it is 0

Answer 5

See, if x->0
IT means its inverse tends to infinity. That is e^(1/x) tends to infinity.
Therefore the constant added ( +1 and -1) can be ignored in both numerator and denominator.
Hence The fraction simply is 1.

Answer 6

What am I missing?

Answer 7

Doesn't it depend on the direction \(x\to0\) ?

Answer 8

note : the limit not allowed be 0/0 or ~/~

Answer 9

pellet. It has to depend on the direction. Absolutely, @SithsAndGiggles

Answer 10

I think it should be not defined then.
I dont see how its 0.

Answer 11

i think limit does not exist

Answer 12

Yeah. Exactly.

Answer 13

if directly we subtitute x = 0
it is (e^~ - 1)/(e^ + 1) = ~/~ , right ?

Answer 14

Using your substitution,\[x\to0^+~\Rightarrow~t\to+\infty\\
x\to0^-~\Rightarrow~t\to-\infty\]