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

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

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

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

1

Solution steps

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

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

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

Simplify

ln (x + 1) \frac{1}{x} : \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 cos^{2} (10 x) d x

\frac{d}{d x} ((x^{4} + 2)^{c o s (x)})

x \to - 3 lim (x^{3})

y^{^{''}} + 9 y = 2 sec (3 t)

\frac{d}{d x} (\frac{i}{x})

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