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

1) Compute the limit of

x \cdot \ln (x)

x \to 0

⁺

ⓘ This is a reference limit of the function ln

The limit of

\ln (x)

-∞

when

x \to 0

⁺

For

x \cdot \ln (x)

, we encounter the indeterminate form

0 \cdot -∞

x

tends to

0

2) To resolve the indeterminate form 0 × ∞, we rewrite the product

x \cdot \ln (x)

as the quotient

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

ⓘ A nonzero number divided by a quantity that tends to 0 grows without bound, the signs giving +∞ or -∞

The limit of

\frac{1}{x}

+∞

when

x \to 0

⁺

For

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

, the limit takes the indeterminate form

- \frac{+∞}{+∞}

when

x

→

0

3) 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:

4) Let's differentiate:

\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}

5) Let's perform the differentiation:

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

For

p1

1

and

q2

x

, use the differentiation rule:

\frac{d}{d x} (\frac{p1}{q2}) = \frac{\frac{d}{d x} (p1) \cdot q2 - p1 \cdot \frac{d}{d x} (q2)}{{q2}^{2}}

6) Let's differentiate

p1

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

ⓘ The derivative of a constant is zero:

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

0

0

7) Let's perform the differentiation

q2

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

ⓘ The derivative of x is 1:

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

1

1

8) Let's evaluate the resulting derivative (

\frac{p1}{q2}

)':

\frac{0 \cdot x - 1 \cdot 1}{x^{2}}

ⓘ A product with a zero factor is zero:

0 \cdot x

0

■Multiplications are done before additions and subtractions:

1 \cdot 1

1

\frac{0 - 1}{x^{2}}

ⓘ Adding 0 changes nothing:

0 - 1

-1

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

9) Then you can continue the derivation:

ⅰ

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

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

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

Using those derivatives, we obtain a new fraction whose limit can be computed:

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

ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

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

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

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

ⓘ Dividing is multiplying by the reciprocal:

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

- \frac{1}{x} \cdot x^{2}

- \frac{1}{x} \cdot x^{2}

ⓘ Multiplying a fraction by a term multiplies its numerator:

- \frac{1}{x} \cdot x^{2}

- \frac{1 \cdot x^{2}}{x}

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

■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor:

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

- x

- x

The limit of

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

0

⁻ when

x \to 0

⁺

So the limit of

x \cdot \ln (x)

0

⁻ when

x \to 0

⁺

▶Limit:

0