What is the limit as x approaches 0 of (1+x)^{1/(3x)} ?

The limit as x approaches 0 of (1+x)^{1/(3x)} is cube root of e

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limit as x approaches 0 of (1+x)^{1/(3x)}

x \to 0 lim ((1 + x)^{\frac{1}{3 x}})

3 e

Decimal

1.39561 \dots

Solution steps

x \to 0 lim ((1 + x)^{\frac{1}{3 x}})

Apply exponent rules

= x \to 0 lim (e^{\frac{1}{3 x} l n (1 + x)})

Apply the Limit Chain Rule:

3 e

= 3 e

y = 3 x + 3 x

\frac{d y}{d x} + y = e^{x} y^{- 2}

y = x^{3}, at x = - 4

\int_{- 4}^{0} (4 + 16 - x^{2}) d x

\int - 5 e^{6 x} d x

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