limit as x approaches 0+of x^{20x} — step-by-step solution

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

1) Compute the limit of

x^{(20 x)}

x \to 0

⁺

The limit of

20 x

is (

20 \cdot 0

)⁺ when

x \to 0

⁺

⇥1.1) Let's evaluate:

20 \cdot 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)

The limit of

\ln (x)

-∞

when

x \to 0

⁺

For

20 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

20 x \cdot \ln (x)

as the quotient

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

The limit of

\frac{1}{20 x}

+∞

when

x \to 0

⁺

For

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

Let's perform the differentiation:

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

\frac{1}{x}

4) Let's differentiate:

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

We differentiate

u1

1

and

v2

20 x

using:

\frac{d}{d x} (\frac{u1}{v2}) = \frac{\frac{d}{d x} (u1) \cdot v2 - u1 \cdot \frac{d}{d x} (v2)}{{v2}^{2}}

5) Let's perform the differentiation

u1

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

0

6) Let's differentiate

v2

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

20

7) You can evaluate the resulted derivative (

\frac{u1}{v2}

)':

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

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

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

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

- \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}}

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

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

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

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

ℹ

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

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

‖

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

-20 x

-20 x

The limit of

- 20 x

is (

- 20 \cdot 0

)⁻ when

x \to 0

⁺

⇥7.1) Let's evaluate:

- 20 \cdot 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