What is the limit as x approaches 0 of (1+4x)^{3/x} ?

The limit as x approaches 0 of (1+4x)^{3/x} is e^{12}

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limit as x approaches 0 of (1+4x)^{3/x}

x \to 0 lim ((1 + 4 x)^{\frac{3}{x}})

e^{12}

Decimal

162754.79141 \dots

Solution steps

x \to 0 lim ((1 + 4 x)^{\frac{3}{x}})

Apply exponent rules

= x \to 0 lim (e^{\frac{3}{x} l n (1 + 4 x)})

Apply the Limit Chain Rule:

e^{12}

= e^{12}

\int (\frac{tan ( u )}{4} - \frac{4}{7} u^{6} - \frac{2}{u}) d u

x \to a lim (\frac{1}{2})

d er i v a t i v e f (x) = (5 x + 4)^{3}

\int_{0}^{5} \frac{1}{u ^{2} - 25} d u

t an g e n t y = 4 cos (x), (\frac{π}{3}, 2)

What is the limit as x approaches 0 of (1+4x)^{3/x} ?
The limit as x approaches 0 of (1+4x)^{3/x} is e^{12}