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

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

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

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

1

Solution steps

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

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

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= 1

\int (3 x - 2)^{4} d x

x^{2}, 0

f (x) = \frac{( 1 - 4 x )}{1 + 4 x}

t \frac{d y}{d t} = 4 y

n = 0 \sum \infty (\frac{e}{8 π})^{n}

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