limit x->0 ln(1+x)/x — step-by-step solution

1) Compute the limit of

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

x \to 0

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

The limit of

1 + x

1

when

x \to 0

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

The limit of

\ln (1 + x)

\ln (1)

when

x \to 0

Let's calculate:

\ln (1)

ⓘ ln(1) = 0, since exp(0) = 1:

\ln (1)

0

0

For

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

, the limit takes the indeterminate form 0/0 when

x

→

0

2) We can use L'Hôpital's rule (limit of a/b = limit of a'/b') by computing the derivatives of the numerator and the denominator:

3) Let's differentiate:

\frac{d}{d x} (\ln (1 + x))

Apply the differentiation rule:

\frac{d}{d x} (\ln (f)) = \frac{d}{d x} (f) \cdot \frac{1}{f}

with f =

1 + x

ⓘ The derivative of ln(x) is 1/x:

\frac{d}{d x} (\ln (1 + x))

\frac{d}{d x} (1 + x) \cdot \frac{1}{1 + x}

\frac{d}{d x} (1 + x) \cdot \frac{1}{1 + x}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{d}{d x} (1 + x) \cdot \frac{1}{1 + x}

\frac{\frac{d}{d x} (1 + x) \cdot 1}{1 + x}

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

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

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

0 + 1

\frac{0 + 1}{1 + x}

ⓘ Adding 0 changes nothing:

0 + 1

1

\frac{1}{1 + x}

4) Let's differentiate:

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

ⓘ The derivative of x is 1:

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

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}{1 + x}

1

when

x \to 0

So the limit of

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

1

when

x \to 0

▶Limit:

1