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

The limit as x approaches 0+of (e^x)/(1-e^{-x)} is infinity

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limit as x approaches 0+of (e^x)/(1-e^{-x)}

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

\infty

Solution steps

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

Factor out

e^{- x}

from

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

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

Simplify

\frac{e ^{x}}{e ^{- x} ( \frac{- 1}{- e ^{- x}} - 1 )} : \frac{e ^{2 x}}{e ^{x} - 1}

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

\frac{e ^{2 x}}{e ^{x} - 1} = e^{2 x} \frac{1}{e ^{x} - 1}

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

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

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

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

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

= 1 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

\frac{d}{d x} (4 e^{6 x})

f (x) = sin (3 x + 1)

\frac{d}{d x} ((4 x)^{6 x})

x \to π lim (x cos (\frac{7 x}{6}))

\frac{x}{x + 6}

What is the limit as x approaches 0+of (e^x)/(1-e^{-x)} ?
The limit as x approaches 0+of (e^x)/(1-e^{-x)} is infinity