Evaluate the limit of the function (x^x- 1)/(x-1), if x goes to 1 but do not use L'Hospital's theorem.

Let f(x) = x^x

If we'll put x
= 1 => f(1) = 1^1  =1

We'll re-write the function
whose limit has to be found out:

lim (f(x) - 1)/(x -
1)

By definition, the derivative of a function f(x), at the
point x = 1 is: lim (f(x) - 1)/(x - 1) = f'(1).

We'll have
to determine the expression of the 1st derivative of
f(x).

We'll take natural logarithms both
sides:

ln f(x) = ln (x^x)

ln
f(x) = x*ln x

We'll differentiate with respect to x both
sides:

f'(x)/f(x) = ln x +
x/x

f'(x)/f(x) = ln x +
1

f'(x) = f(x)*(ln x +
1)

f'(x) = (x^x)*(ln x +
1)

Now, we'll replace x by
1:

f'(1) = 1*(ln 1 + 1)

f'(1)
= 1

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