Question

Examine whether lim_"x→0"e^(1/x)/e^(1/x)+1 exits or not?

Answer 1

The limit does not exist.

Explanation:

Assume for the moment that x approaches 0 from the positive side.
Then we want
`lim_(x -> 0+) e^(1/x)/(e^(1/x) + 1)`.
Let `y = 1/x`, and the above equals
`lim_(y -> oo) e^y/(e^y + 1)`.
The numerator and denominator approach infinity. Applying L'Hopital's Rule, take the derivatives of the numerator and denominator. Then the above equals:
`lim_(y -> oo) e^y/e^y = lim_(y -> oo)1 = 1`.

Now assume that x approaches 0 from the negative side.
Then
`lim_(x -> 0-) e^(1/x)/(e^(1/x) + 1)`.
`=lim_(y -> -oo) e^y/(e^y + 1)`.
The exponentials go to zero, and the denominator goes to 1.
`= 0/(0 + 1) = 0`.

Since the left- and right- and limits are not equal, the limit does not exist.