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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Simplify

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

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

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

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{1}{2} \cdot \frac{4 e ^{2 \cdot 0}}{4 e ^{2 \cdot 0} \cdot 0 + 4 e ^{2 \cdot 0}}

Simplify

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

= \frac{1}{2}

\frac{\partial}{\partial x} (x^{2})

7 x - 2 x

y = 20 x^{3} (x^{4} + 5)^{4}

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

\frac{\partial}{\partial x} (- sin (y))

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