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

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

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limit as x approaches 0+of 2xln(x)

x \to 0 + lim (2 x ln (x))

0

Solution steps

x \to 0 + lim (2 x ln (x))

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

= 2 \cdot x \to 0 + lim (x ln (x))

Rewrite for L'Hopital

= 2 \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}}})

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

Simplify

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

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

Plug in the value

x = 0

= 2 (- 0)

Simplify

= 0

x \to 0 lim (\frac{( x - 3 ) ^{2} - 9}{x})

x \to 0 + lim ((tan (x))^{s i n (x)})

x \to 6 lim (\frac{2 x}{x ^{2} - 36})

x \to 0 + lim ((1 - 3 x)^{\frac{2}{x}})

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

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