What is the limit as x approaches 0+of 2x(ln(x))^2 ?

The limit as x approaches 0+of 2x(ln(x))^2 is 0

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limit as x approaches 0+of 2x(ln(x))^2

x \to 0 + lim (2 x (ln (x))^{2})

0

Solution steps

x \to 0 + lim (2 x (ln (x))^{2})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 2 \cdot x \to 0 + lim (x (ln (x))^{2})

Rewrite for L'Hopital

= 2 \cdot x \to 0 + lim (\frac{( ln ( x ) ) ^{2}}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}})

= 2 \cdot x \to 0 + lim (\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}})

Simplify

\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}} : - 2 x ln (x)

= 2 \cdot x \to 0 + lim (- 2 x ln (x))

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 2 (- 2 \cdot x \to 0 + lim (x ln (x)))

Rewrite for L'Hopital

= 2 (- 2 \cdot x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x}}))

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}})

= 2 (- 2 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}}))

Simplify

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

= 2 (- 2 \cdot x \to 0 + lim (- x))

Plug in the value

x = 0

= 2 (- 2 (- 0))

Simplify

= 0

x \to 1 lim (x^{\frac{2}{1 - x ^{2}}})

x \to - 2 - lim (\frac{∣ x + 2 ∣}{x + 2})

x \to 1 lim (\frac{ln ( x )}{e ^{x}})

x \to \infty lim (3 x^{2} + x)

x \to - \infty lim (- 3 x^{4})

What is the limit as x approaches 0+of 2x(ln(x))^2 ?
The limit as x approaches 0+of 2x(ln(x))^2 is 0