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

f (x) = x^{2} + 9, (5, 34)

\int \frac{( 9 x ^{2} + 5 x + 9 )}{( x ^{2} + 1 ) ^{2}} d x

(x^{6} + y^{6}) d x + 6 x y^{5} d y = 0

f (t) = cos^{2} (t)

\frac{d}{d x} (ln (1 - 4 x))

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