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

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

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

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

2

Solution steps

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 2

\frac{\partial}{\partial x} ((6 x + 3 y) ey)

\int \frac{1}{( x - 3 ) ( x + 2 )} d x

f (x) = x^{2} + 6, (4, 22)

(1, 6), (0.5, 8)

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

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