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

The limit as x approaches 0 of (1+4x)^{4/x} is e^{16}

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

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

e^{16}

Solution steps

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

Apply exponent rules

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

Apply the Limit Chain Rule:

e^{16}

= e^{16}

\frac{\partial}{\partial x} \frac{2 t}{( 1 - \frac{0.5 ( x )}{L} ) ^{2}}

\frac{d}{d x} (\frac{1}{csc ( x ) + cot ( x )})

2 arcsin (x^{3})

\frac{d}{d x} (10 e^{- 2 x})

h (t) = (t + 5)^{\frac{2}{3}} (3 t^{2} - 5)^{3}

What is the limit as x approaches 0 of (1+4x)^{4/x} ?
The limit as x approaches 0 of (1+4x)^{4/x} is e^{16}