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

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

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

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

diverges

Solution steps

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

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

then the limit does not exist

x \to 1 + lim (\frac{e ^{x - 1} - 1}{ln ^{2} ( x )}) = \infty

x \to 1 - lim (\frac{e ^{x - 1} - 1}{ln ^{2} ( x )}) = - \infty

= diverges

x^{2} - 4 x + 3, 3 + 4 x - x^{2}

\int_{- 1}^{1} (x^{2} - 1) d x

\frac{\partial}{\partial z} (sin (x + 2 y + 3 z))

f (x) = 6 x^{2} (x - 9)

\frac{\partial}{\partial x} (\frac{2}{x ^{3}})

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