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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

= x \to 0 + lim (x)

Plug in the value

x = 0

= 0

t \to 0 lim (\frac{1}{t} - \frac{1}{t ^{2} - t})

x \to 0 lim (\frac{sin ( a ) x}{x})

x \to 0 + lim (arcsin (x - 1))

x \to 0 lim (x^{3} + \frac{cos ( 5 x )}{10000})

x \to 0 + lim (9 x ln (x))

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