What is the limit as x approaches 0+of xln(1/(x^2)) ?

The limit as x approaches 0+of xln(1/(x^2)) is 0

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limit as x approaches 0+of xln(1/(x^2))

x \to 0 + lim (x ln (\frac{1}{x ^{2}}))

0

Solution steps

x \to 0 + lim (x ln (\frac{1}{x ^{2}}))

Simplify

x ln (\frac{1}{x ^{2}}) : - x ln (x^{2})

= x \to 0 + lim (- x ln (x^{2}))

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

= - x \to 0 + lim (x ln (x^{2}))

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

= - x \to 0 + lim (- 2 x)

Plug in the value

x = 0

= - (- 2 \cdot 0)

Simplify

= 0

x \to 2 lim (3 + x - 2)

x \to 2 + lim (\frac{x + 2}{x - 2})

x \to \frac{1}{3} lim (ln (x))

x \to 2 lim (\frac{1}{x + 3} + 5)

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

What is the limit as x approaches 0+of xln(1/(x^2)) ?
The limit as x approaches 0+of xln(1/(x^2)) is 0