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

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

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

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

1

Solution steps

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

x \to 0 lim (\frac{6}{x ^{2} + x} - \frac{6}{x})

\int_{- 1}^{3} x (x - 1) (x + 2) d x

\int x^{3} x^{2} + 19 d x

y = \frac{4}{5 x + 1}, (- 1, - 1)

\frac{9}{x ^{2}} - \frac{3}{x}

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