What is the limit as x approaches 0 of x/(ln(2x+1)) ?

The limit as x approaches 0 of x/(ln(2x+1)) is 1/2

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 x/(ln(2x+1))

x \to 0 lim (\frac{x}{ln ( 2 x + 1 )})

\frac{1}{2}

Decimal

0.5

Solution steps

x \to 0 lim (\frac{x}{ln ( 2 x + 1 )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{1}{\frac{2}{2 x + 1}})

= x \to 0 lim (\frac{1}{\frac{2}{2 x + 1}})

Plug in the value

x = 0

= \frac{1}{\frac{2}{2 \cdot 0 + 1}}

Simplify

= \frac{1}{2}

x \to \infty lim ((1 + \frac{0.02}{x})^{2 x})

x \to 0 lim (3 x^{2} + 2)

x \to \infty lim ((\frac{2 x + 1}{x + 3})^{\frac{3 x + 2}{x - 5}})

x \to 0 lim (\frac{sin ( x )}{arcsin ( x )})

x \to - \infty lim (x + (x))

What is the limit as x approaches 0 of x/(ln(2x+1)) ?
The limit as x approaches 0 of x/(ln(2x+1)) is 1/2