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

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

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

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

\infty

Solution steps

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

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

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

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

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

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

With the exception of indeterminate form

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

t \to 0 + lim (\frac{1}{t}) = \infty

t \to 0 + lim (1 - \frac{\frac{1}{e ^{t} - 1}}{\frac{4}{t}}) = \frac{3}{4}

= 4 \cdot \infty \cdot \frac{3}{4}

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

\frac{d}{d x} (2 csc (x) cot (x))

y = 2 x + 1, at x = \frac{1}{4}

\frac{s ^{2}}{( s - 3 ) ( s ^{2} + 5 )}

n = 0 \sum \infty \frac{e ^{n}}{n ^{2}}

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

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