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

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

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limit as x approaches 0 of 1/x*ln(x+1)

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

1

Solution steps

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

Simplify

\frac{1}{x} ln (x + 1) : \frac{ln ( x + 1 )}{x}

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= 1

\int 6 x yz + e^{x} d y

\frac{\partial}{\partial y} (4 x + 8 y - 88)

x \to 0 + lim ((x) ln (x^{4}))

\frac{40}{1 + 25 p + 100 p ^{2}}

\frac{\partial}{\partial x} (x yz - sin (x + y + z))

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