The limit `lim_(x->1) (e^x -e)/(x-1)` , has to be determined.

If we substitute x = 1, the value of the numerator is e^1 - e = 0 and the value of the denominator is also 1 - 1 = 0. The final result is `0/0` which is indeterminate and in a form that allows the use of l'Hospital's rule to substitute the numerator and denominator by their derivatives.

The given limit can now be written as:

`lim_(x->1) e^x/1`

Substituting x = 1 gives e^1 = e

The required limit `lim_(x->1) (e^x -e)/(x-1) = e`

We notice that the expression lim (e^x -e)/(x-1) is the way to determine the derivative of the function f(x) = e^x, at the point x = 1, using the first principle.

lim [f(x) - f(1)]/(x-1) = lim (e^x - e)/(x-1) = f'(1)

Let f(x) = e^x.

We'll take logarithms both sides:

ln f(x) = ln e^x

We'll apply the power rule of logarithms:

ln f(x) = x*ln e, but ln e = 1

ln f(x) = x

We'll differentiate both sides with respect to x:

f'(x)/f(x) = 1 => f'(x) = f(x) = e^x

f'(1) = e^1 =e

**Therefore, the limit of the function, if x approaches to 1, is lim (e^x -e)/(x-1) = e.**