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

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

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

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

\frac{3}{2}

Decimal Notation

1.5

Solution steps

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

t = 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}

y^{^{'}} + y = t e^{- t} + 1

\int \frac{1}{12} d x

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

\frac{\partial}{\partial x} (2 e^{x - 3 y})

x^{4} + y^{4}

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