What is the limit as x approaches 0+of 1/(xln(x^9)) ?

The limit as x approaches 0+of 1/(xln(x^9)) is -infinity

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limit as x approaches 0+of 1/(xln(x^9))

x \to 0 + lim (\frac{1}{x ln ( x ^{9} )})

- \infty

Solution steps

x \to 0 + lim (\frac{1}{x ln ( x ^{9} )})

ln (x^{9}) = 9 ln (x)

= x \to 0 + lim (\frac{1}{x \cdot 9 ln ( x )})

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

= \frac{1}{9} \cdot x \to 0 + lim (\frac{1}{x ln ( x )})

\frac{1}{x ln ( x )} = \frac{1}{x} \cdot \frac{1}{ln ( x )}

= \frac{1}{9} \cdot x \to 0 + lim (\frac{1}{x} \cdot \frac{1}{ln ( x )})

Rewrite for L'Hopital

= \frac{1}{9} \cdot x \to 0 + lim \frac{\frac{1}{x}}{\frac{1}{\frac{1}{l n ( x )}}}

Apply L'Hopital's Rule

= \frac{1}{9} \cdot x \to 0 + lim (\frac{- \frac{1}{x ^{2}}}{\frac{1}{x}})

Simplify

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

= \frac{1}{9} \cdot x \to 0 + lim (- \frac{1}{x})

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

= \frac{1}{9} (- x \to 0 + lim (\frac{1}{x}))

For

x

approaching

0

from the right

, x > 0 \Rightarrow x > 0

The denominator is a positive quantity approaching 0 from the right

= \frac{1}{9} (- \infty)

Apply Infinity Property:

c \cdot (- \infty) = - \infty

= - \infty

What is the limit as x approaches 0+of 1/(xln(x^9)) ?
The limit as x approaches 0+of 1/(xln(x^9)) is -infinity