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

The limit as x approaches 0 of 1/(2^{1/x)-1} is diverges

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limit as x approaches 0 of 1/(2^{1/x)-1}

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

diverges

Solution steps

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

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

= diverges

\frac{d}{d x} (sin (5 x - 2))

\int in (x + 1) d x

\int \frac{1}{z - 1} d z

(x^{3} + 3) (3 x^{2} - 4 x + 2)

x \to 6 lim (\frac{x + 2}{( x + 6 )})

What is the limit as x approaches 0 of 1/(2^{1/x)-1} ?
The limit as x approaches 0 of 1/(2^{1/x)-1} is diverges