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

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

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

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

diverges

Solution steps

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

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

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

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

= diverges

x \to 4 lim (2 + 4 x)

x \to 3 lim (\frac{2 x + 1}{x - 1})

x \to \frac{π}{6} lim (2 sin (x))

x \to 2 lim (3 x^{3} + 7 x - 16)

z \to 0 lim (cot (z) - csc (z))

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