What is the limit as x approaches 1 of 4/(x^3-1) ?

The limit as x approaches 1 of 4/(x^3-1) is diverges

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limit as x approaches 1 of 4/(x^3-1)

x \to 1 lim (\frac{4}{x ^{3} - 1})

diverges

Solution steps

x \to 1 lim (\frac{4}{x ^{3} - 1})

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

then the limit does not exist

x \to 1 + lim (\frac{4}{x ^{3} - 1}) = \infty

x \to 1 - lim (\frac{4}{x ^{3} - 1}) = - \infty

= diverges

ln (\frac{6 - x}{6 + x})

\int 2 - y^{2} d y

\frac{d}{d x} (\frac{1}{8} (x^{2} - 8 ln (x)))

\int \frac{e ^{x}}{( e ^{x} - 3 ) ( e ^{2 x} + 1 )} d x

\frac{d}{d x} (e^{x^{2} - y})

What is the limit as x approaches 1 of 4/(x^3-1) ?
The limit as x approaches 1 of 4/(x^3-1) is diverges