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

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

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limit as x approaches 0+of xln(x)

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

0

Solution steps

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

Rewrite for L'Hopital

= 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}}})

= 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

Simplify

= 0

\frac{d}{d x} (sin (6 x) cos (4 x))

x^{4} + y^{3} x^{5} + 5 y^{^{'}} = 0

3 x^{2} y^{2} d x + 2 x^{3} y d y = 0

f (x) = 2 - 4 x

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