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

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

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

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

2

Solution steps

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

t = 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

x \to 4 lim (2 x - 1)

\frac{d}{d x} (5 e^{4 x})

\int \frac{sin ^{5} ( x )}{x} d x

\int x^{2} 3 7 x^{3} + 1 d x

x \to - \infty lim (4 e^{- x})

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