What is the limit as x approaches 0+of xln(x^8) ?

The limit as x approaches 0+of xln(x^8) is 0

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limit as x approaches 0+of xln(x^8)

x \to 0 + lim (x ln (x^{8}))

0

Solution steps

x \to 0 + lim (x ln (x^{8}))

ln (x^{8}) = 8 ln (x)

= x \to 0 + lim (x \cdot 8 ln (x))

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

= 8 \cdot x \to 0 + lim (x ln (x))

Rewrite for L'Hopital

= 8 \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}}})

= 8 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}})

Simplify

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

= 8 \cdot x \to 0 + lim (- x)

Plug in the value

x = 0

= 8 (- 0)

Simplify

= 0

y^{^{'}} = (0.04 \cdot 4) - \frac{y}{1840} 4

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

\frac{\partial}{\partial y} (sin (x^{3} + y^{2}))

\frac{d}{d x} (\frac{3 x + 1}{x + 1})

\frac{2}{s ^{2} - 2 s + 9}

What is the limit as x approaches 0+of xln(x^8) ?
The limit as x approaches 0+of xln(x^8) is 0