What is the limit as x approaches 0+of x^{6/(1+2ln(x))} ?

The limit as x approaches 0+of x^{6/(1+2ln(x))} is e^3

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limit as x approaches 0+of x^{6/(1+2ln(x))}

x \to 0 + lim (x^{\frac{6}{1 + 2 l n ( x )}})

e^{3}

Decimal Notation

20.08553 \dots

Solution steps

x \to 0 + lim (x^{\frac{6}{1 + 2 l n ( x )}})

Apply exponent rules

= x \to 0 + lim (e^{\frac{6}{1 + 2 l n ( x )} l n (x)})

Apply the Limit Chain Rule:

e^{3}

= e^{3}

x \to 0 - lim (3 e^{- \frac{9}{x}})

x \to 0 lim (1 x^{2})

x \to 4 + lim (2 + x - 4)

x \to 8 lim (7 - \frac{1}{2} x)

x \to 0 lim ((1 - x)^{2})

What is the limit as x approaches 0+of x^{6/(1+2ln(x))} ?
The limit as x approaches 0+of x^{6/(1+2ln(x))} is e^3