What is the limit as x approaches 0+of-4xln(5x) ?

The limit as x approaches 0+of-4xln(5x) is 0

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limit as x approaches 0+of-4xln(5x)

x \to 0 + lim (- 4 x ln (5 x))

0

Solution steps

x \to 0 + lim (- 4 x ln (5 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 (5 x))

Rewrite for L'Hopital

= - 4 \cdot x \to 0 + lim (\frac{ln ( 5 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 0 lim (\frac{x ^{3} + 4}{x ^{2}})

x \to 0 lim (\frac{3 x + 4}{x})

x \to π lim (cos (9 x))

x \to - π lim (ln (sin (x)))

j \to 0 lim (\frac{( 3 - j ) ^{2} - 9}{j})

What is the limit as x approaches 0+of-4xln(5x) ?
The limit as x approaches 0+of-4xln(5x) is 0