What is the limit as x approaches 0+of (x+1)^{ln(x)} ?

The limit as x approaches 0+of (x+1)^{ln(x)} is 1

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limit as x approaches 0+of (x+1)^{ln(x)}

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

1

Solution steps

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

Apply exponent rules

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

Apply the Limit Chain Rule:

1

= 1

\frac{d}{d x} (ln (e^{4 x + 1}))

\frac{\partial}{\partial x} (\frac{5 x y}{10 x - y})

x \to \frac{π}{2} lim (x sec (x))

\frac{4}{s ( s + 2 ) ^{3}}

\int e^{l n (t)} d t

What is the limit as x approaches 0+of (x+1)^{ln(x)} ?
The limit as x approaches 0+of (x+1)^{ln(x)} is 1