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

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

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

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

\infty

Solution steps

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

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

With the exception of indeterminate form

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

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

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

= \infty - (- \infty)

Apply Infinity Property:

\infty + \infty = \infty

= \infty

x \to - 3 lim (4^{x})

x \to - \infty lim (- 3 x^{5} + 2 x - 4)

x \to 3 lim (\frac{x ^{2} + 3}{x + 3})

x \to 1 lim (x^{l n (x)})

x \to 2 lim (\frac{sin ( π x )}{x ^{2} - 4})

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