limit x->0+ x^{20x} — step-by-step solution

⚠ x^{20x} is read as x^(20x): use ^(...) to group an exponent, use * to multiply

1) Let's evaluate the limit of

x^{20 x}

when

x \to 0

⁺

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

The limit of

20 x

is (

20 \cdot 0

)⁺ when

x \to 0

⁺

Let's evaluate:

20 \cdot 0

ⓘ A product with a zero factor is zero:

20 \cdot 0

0

0

For

x^{20 x}

, the limit takes the indeterminate form

0^{0}

when

x

→

0

For a positive base, rewrite

x^{20 x}

\exp (20 x \cdot \ln (x))

, then compute the limit of

20 x \cdot \ln (x)

ⓘ This is a reference limit of the function ln

The limit of

\ln (x)

-∞

when

x \to 0

⁺

For

20 x \cdot \ln (x)

, the limit takes the indeterminate form

0 \cdot -∞

when

x

→

0

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

20 x \cdot \ln (x)

as the quotient

\frac{\ln (x)}{\frac{1}{20 x}}

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

The limit of

\frac{1}{20 x}

+∞

when

x \to 0

⁺

For

\frac{\ln (x)}{\frac{1}{20 x}}

, we encounter the indeterminate form

- \frac{+∞}{+∞}

x

tends to

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 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}

5) Let's perform the differentiation:

\frac{d}{d x} (\frac{1}{20 x})

For

p1

1

and

q2

20 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 differentiate

q2

\frac{d}{d x} (20 x)

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (20 x)

20

20

8) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

\frac{0 \cdot 20 x - 1 \cdot 20}{{(20 x)}^{2}}

ⓘ A product with a zero factor is zero:

0 \cdot 20 x

0

\frac{0 - 20}{{(20 x)}^{2}}

ⓘ Adding 0 changes nothing:

0 - 20

-20

- \frac{20}{{(20 x)}^{2}}

ⓘ The power of a product is the product of the powers:

{(20 x)}^{2}

20^{2} \cdot x^{2}

- \frac{20}{20^{2} \cdot x^{2}}

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

20^{2}

20 \cdot 20

400

- \frac{20}{400 x^{2}}

We simplify the fraction:

\frac{20}{400}

by the common factor:

20

to get the new fraction:

\frac{1}{20}

- \frac{1}{20 x^{2}}

9) Then you can continue the derivation:

ⅰ

\frac{d}{d x} (\frac{1}{20 x})

- \frac{1}{20 x^{2}}

- \frac{1}{20 x^{2}}

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

\frac{\frac{1}{x}}{- \frac{1}{20 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}{20 x^{2}}}

- \frac{\frac{1}{x}}{\frac{1}{20 x^{2}}}

- \frac{\frac{1}{x}}{\frac{1}{20 x^{2}}}

ⓘ Dividing is multiplying by the reciprocal:

- \frac{\frac{1}{x}}{\frac{1}{20 x^{2}}}

- \frac{1}{x} \cdot 20 x^{2}

- \frac{1}{x} \cdot 20 x^{2}

ⓘ Multiplying a fraction by a term multiplies its numerator:

- \frac{1}{x} \cdot 20 x^{2}

- \frac{1 \cdot 20 x^{2}}{x}

- \frac{20 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{20 x^{2}}{x}

-20 x

-20 x

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

The limit of

- 20 x

is (

- 20 \cdot 0

)⁻ when

x \to 0

⁺

Let's calculate:

- 20 \cdot 0

ⓘ A product with a zero factor is zero:

20 \cdot 0

0

0

The limit of

\frac{\ln (x)}{\frac{1}{20 x}}

0

⁻ when

x \to 0

⁺

The limit of

20 x \cdot \ln (x)

0

⁻ when

x \to 0

⁺

So the limit of

x^{20 x}

1

⁻ when

x \to 0

⁺

▶Limit:

1