What is the limit as x approaches 0 of 1/(ln(x+1)) ?

The limit as x approaches 0 of 1/(ln(x+1)) is diverges

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limit as x approaches 0 of 1/(ln(x+1))

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

diverges

Solution steps

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

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

= diverges

\int cos (\frac{x}{y}) d x

\frac{d}{d x} (\frac{( e ^{- x} )}{x})

f (x) = \frac{x - 2}{x ^{2} - 2 x - 3}

f (x) = - \frac{3}{2} x^{2} + 3 x + 3

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

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