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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:
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.
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