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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

sin (3 x) + cos (8 x)

(1 + x^{2}) (\frac{d y}{d x}) - 2 x y = 0

\int_{\frac{π}{2}}^{π} x - sin (x) d x

\frac{d y}{d x} = \frac{3 x}{4 y}

sec (3 sin (2 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