What is the limit as x approaches 0+of 4xln(x) ?

The limit as x approaches 0+of 4xln(x) is 0

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limit as x approaches 0+of 4xln(x)

x \to 0 + lim (4 x ln (x))

0

Solution steps

x \to 0 + lim (4 x ln (x))

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

= 4 \cdot x \to 0 + lim (x ln (x))

Rewrite for L'Hopital

= 4 \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}}})

= 4 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}})

Simplify

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

= 4 \cdot x \to 0 + lim (- x)

Plug in the value

x = 0

= 4 (- 0)

Simplify

= 0

x \to - \infty lim (\frac{6}{x})

x \to 0 lim (3 x cot (2 x))

x \to 0 - lim (\frac{2}{x ^{3}})

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

x \to 0 + lim (x sin (ln (x)))

What is the limit as x approaches 0+of 4xln(x) ?
The limit as x approaches 0+of 4xln(x) is 0