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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

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

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

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

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

x \to 2 lim (\frac{x}{( x ^{3} - 8 )})

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