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

The limit as x approaches 0+of xln(e^x-1) is 0

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limit as x approaches 0+of xln(e^x-1)

x \to 0 + lim (x ln (e^{x} - 1))

0

Solution steps

x \to 0 + lim (x ln (e^{x} - 1))

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{- e ^{0} \cdot 0 ^{2} - 2 e ^{0} \cdot 0}{e ^{0}}

Simplify

\frac{- e ^{0} \cdot 0 ^{2} - 2 e ^{0} \cdot 0}{e ^{0}} : 0

= 0

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