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

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

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

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

0

Solution steps

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

ln (x^{4}) = 4 ln (x)

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

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

\int_{- 2}^{3} \frac{26}{x ^{4}} d x

\frac{d}{d x} (\frac{1}{1 + x ^{4}})

\frac{d}{d x} (4 sec (x) - 8 cos (x))

y = x^{3}, y = x, x \geq 0

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