What is the limit as x approaches 0+of (e^x-1)^x ?

The limit as x approaches 0+of (e^x-1)^x is 1

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limit as x approaches 0+of (e^x-1)^x

x \to 0 + lim ((e^{x} - 1)^{x})

1

Solution steps

x \to 0 + lim ((e^{x} - 1)^{x})

Apply exponent rules

= x \to 0 + lim (e^{x l n (e^{x} - 1)})

Apply the Limit Chain Rule:

1

= 1

\int t (ln (t))^{2} d t

x \to 0 - lim (\frac{5}{tan ( x )})

\int e^{- x} \cdot x^{2} d x

arcsin (\frac{1}{t})

\frac{\partial}{\partial x} (e^{- 7 x} cos (3 π t))

What is the limit as x approaches 0+of (e^x-1)^x ?
The limit as x approaches 0+of (e^x-1)^x is 1