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 e

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

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

e

Decimal Notation

2.71828 \dots

Solution steps

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

Apply exponent rules

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

e^{\frac{1}{l n ( x )} l n (x)} = e

= x \to 0 + lim (e)

x \to a lim c = c

= e

y = x^{4} + 6 e^{x}, (0, 6)

5 \cdot t \cdot e^{- 5 \cdot t}

\frac{\partial}{\partial y} (x^{2} y + x)

\frac{d}{d x} (x^{2} (x - 3))

f (y) = tan (y)

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 e