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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

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

Plug in the value

x = 0

= - \frac{0}{0 + 1}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

x \to 2 lim (2 x^{2} + x - 4)

x \to - 2 lim (\frac{8}{x ^{2} - 4})

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

h (x) \to 0 lim ((x + h (x))^{n})

x \to 2 lim (\frac{x ^{2} + 5 x - 14}{x ^{2} - 11 x + 18})

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