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

The limit as x approaches 1 of (ln(x))/(1-x) is -1

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limit as x approaches 1 of (ln(x))/(1-x)

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

- 1

Solution steps

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

Multiply by the conjugate of

1 - x : \frac{1 - x ^{2}}{1 + x}

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

Simplify

\frac{ln ( x )}{\frac{1 - x ^{2}}{1 + x}} : - \frac{ln ( x )}{x - 1}

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

Apply L'Hopital's Rule

: x \to 1 lim (\frac{\frac{1}{x}}{1})

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

Plug in the value

x = 1

= - \frac{\frac{1}{1}}{1}

Simplify

= - 1

\frac{d}{d x} (16 - x^{2} + 64)

\frac{d x}{d t} = 2 x (7 - x)

\int_{0}^{1} \frac{( x - 1 )}{( x + 1 )} d x

\int 9 x^{2} e^{x} d x

x^{2} = e^{x - y}

What is the limit as x approaches 1 of (ln(x))/(1-x) ?
The limit as x approaches 1 of (ln(x))/(1-x) is -1