What is the limit as x approaches 0 of 3/x-3/(e^x-1) ?

The limit as x approaches 0 of 3/x-3/(e^x-1) is 3/2

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limit as x approaches 0 of 3/x-3/(e^x-1)

x \to 0 lim (\frac{3}{x} - \frac{3}{e ^{x} - 1})

\frac{3}{2}

Decimal Notation

1.5

Solution steps

x \to 0 lim (\frac{3}{x} - \frac{3}{e ^{x} - 1})

Simplify

\frac{3}{x} - \frac{3}{e ^{x} - 1} : \frac{3 ( e ^{x} - 1 ) - 3 x}{x ( e ^{x} - 1 )}

= x \to 0 lim (\frac{3 ( e ^{x} - 1 ) - 3 x}{x ( e ^{x} - 1 )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{3 e ^{x} - 3}{e ^{x} - 1 + e ^{x} x})

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

Apply L'Hopital's Rule

: x \to 0 lim (\frac{3 e ^{x}}{e ^{x} x + 2 e ^{x}})

= x \to 0 lim (\frac{3 e ^{x}}{e ^{x} x + 2 e ^{x}})

Plug in the value

x = 0

= \frac{3 e ^{0}}{e ^{0} \cdot 0 + 2 e ^{0}}

Simplify

\frac{3 e ^{0}}{e ^{0} \cdot 0 + 2 e ^{0}} : \frac{3}{2}

= \frac{3}{2}

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

\int \frac{cos ( x )}{x} d x

\int_{0}^{6} 9 sin (x + 1) d x

- 4 (8 x^{2} + 8)^{- 6}

(1 + x)^{^{'}}

What is the limit as x approaches 0 of 3/x-3/(e^x-1) ?
The limit as x approaches 0 of 3/x-3/(e^x-1) is 3/2