What is the limit as x approaches 0+of (1+2x)^{4/x} ?

The limit as x approaches 0+of (1+2x)^{4/x} is e^8

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limit as x approaches 0+of (1+2x)^{4/x}

x \to 0 + lim ((1 + 2 x)^{\frac{4}{x}})

e^{8}

Decimal Notation

2980.95798 \dots

Solution steps

x \to 0 + lim ((1 + 2 x)^{\frac{4}{x}})

Apply exponent rules

= x \to 0 + lim (e^{\frac{4}{x} l n (1 + 2 x)})

Apply the Limit Chain Rule:

e^{8}

= e^{8}

x^{3} \frac{d ^{3} y}{d x ^{3}} + 2 x^{2} \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} + y = 12 x^{2}

\frac{d ^{3}}{d t ^{3}} ((t^{2} - 1)^{3})

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

\frac{\partial}{\partial x} ((x^{2} y + 4 y) e^{2 x y})

\int \frac{( x ^{2} + 1 )}{5 cosh ^{2} ( x ^{3} + 3 x )} d x

What is the limit as x approaches 0+of (1+2x)^{4/x} ?
The limit as x approaches 0+of (1+2x)^{4/x} is e^8