What is the limit as x approaches 0+of 2/x-1/(e^x-1) ?

The limit as x approaches 0+of 2/x-1/(e^x-1) is infinity

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limit as x approaches 0+of 2/x-1/(e^x-1)

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

\infty

Solution steps

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

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

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

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

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

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

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

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

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

= 2 \cdot \infty \cdot \frac{1}{2}

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

x \to 5 lim (- 2 x + 1)

x \to 0 lim (\frac{a x}{tan ( x )})

x \to 0 + lim (x csc (x))

x \to 0 lim (x \cdot \frac{1}{ln ( x )})

x \to 0 lim (x^{4} cos^{5} (\frac{1}{x}))

What is the limit as x approaches 0+of 2/x-1/(e^x-1) ?
The limit as x approaches 0+of 2/x-1/(e^x-1) is infinity