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

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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.**

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