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

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

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Simplify

\frac{1}{4 x} - \frac{1}{e ^{4 x} - 1} : \frac{( e ^{x} + 1 ) ( e ^{x} - 1 ) ( e ^{2 x} + 1 ) - 4 x}{4 x ( e ^{x} + 1 ) ( e ^{x} - 1 ) ( e ^{2 x} + 1 )}

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

Apply L'Hopital's Rule

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

Apply L'Hopital's Rule

= \frac{1}{4} \cdot x \to 0 lim (\frac{16 e ^{4 x}}{16 e ^{4 x} x + 8 e ^{4 x}})

Plug in the value

x = 0

= \frac{1}{4} \cdot \frac{16 e ^{4 \cdot 0}}{16 e ^{4 \cdot 0} \cdot 0 + 8 e ^{4 \cdot 0}}

Simplify

\frac{1}{4} \cdot \frac{16 e ^{4 \cdot 0}}{16 e ^{4 \cdot 0} \cdot 0 + 8 e ^{4 \cdot 0}} : \frac{1}{2}

= \frac{1}{2}

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