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

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

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

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

\infty

Solution steps

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

Divide by

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

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

For

x

approaching

0

from the right

, x > 0 \Rightarrow 1 - \frac{1}{e ^{x}} > 0

The denominator is a positive quantity approaching 0 from the right

= \infty

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

f (r) = \frac{4}{3} π r^{3}

\int 3 y^{2} z + y e^{x} d x

\int t (t - 1)^{12} d t

\frac{\partial}{\partial x} (sin (5 x - 2 y))

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