What is the limit as x approaches 0+of (-7x)ln(6x^4) ?

The limit as x approaches 0+of (-7x)ln(6x^4) is 0

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limit as x approaches 0+of (-7x)ln(6x^4)

x \to 0 + lim ((- 7 x) ln (6 x^{4}))

0

Solution steps

x \to 0 + lim ((- 7 x) ln (6 x^{4}))

Apply the following algebraic property

: a \cdot b = \frac{a}{\frac{1}{b}}, b \neq = 0

(- 7 x) ln (6 x^{4}) = \frac{ln ( 6 x ^{4} )}{\frac{1}{( - 7 x )}}

= x \to 0 + lim (\frac{ln ( 6 x ^{4} )}{\frac{1}{- 7 x}})

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

= \frac{1}{\frac{1}{- 7}} \cdot x \to 0 + lim (\frac{ln ( 6 x ^{4} )}{\frac{1}{x}})

Simplify

= (- 7) \cdot x \to 0 + lim (\frac{ln ( 6 x ^{4} )}{\frac{1}{x}})

Simplify

= (- 7) \cdot x \to 0 + lim (x ln (6 x^{4}))

Rewrite for L'Hopital

= (- 7) \cdot x \to 0 + lim (\frac{ln ( 6 x ^{4} )}{\frac{1}{x}})

Apply L'Hopital's Rule

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

= (- 7) \cdot x \to 0 + lim (\frac{\frac{4}{x}}{- \frac{1}{x ^{2}}})

Simplify

\frac{\frac{4}{x}}{- \frac{1}{x ^{2}}} : - 4 x

= (- 7) \cdot x \to 0 + lim (- 4 x)

Plug in the value

x = 0

= (- 7) (- 4 \cdot 0)

Simplify

= 0

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What is the limit as x approaches 0+of (-7x)ln(6x^4) ?
The limit as x approaches 0+of (-7x)ln(6x^4) is 0