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

1) Compute the limit of

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

x \to 1

ⓘ The function ln is continuous, its limit is its value at the limit of its argument

The limit of

\ln (x)

\ln (1)

when

x \to 1

⇥1.1) Let's evaluate:

\ln (1)

⇥ⓘ ln(1) = 0, since exp(0) = 1:

\ln (1)

0

⇥

0

ⓘ The limit of a sum of finite limits is the sum of the limits

The limit of

x - 1

1 - 1

when

x \to 1

⇥1.2) Let's calculate:

1 - 1

⇥ⓘ Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

⇥

0

For

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

, we encounter the indeterminate form 0/0 as

x

tends to

1

2) Let's apply L'Hôpital's rule (limit of a/b = limit of a'/b') by differentiating numerator and denominator:

3) Let's perform the differentiation:

\frac{d}{d x} (\ln (x))

ⓘ The derivative of ln(x) is 1/x:

\frac{d}{d x} (\ln (x))

\frac{1}{x}

\frac{1}{x}

4) Let's differentiate:

\frac{d}{d x} (x - 1)

ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (x - 1)

1 + 0

1 + 0

ⓘ Adding 0 changes nothing:

1 + 0

1

1

With the derivatives of the numerator and the denominator, we obtain a new fraction whose limit can be determined:

The limit of

\frac{1}{x}

1

when

x \to 1

So the limit of

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

1

when

x \to 1

▶Limit:

1