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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