What is the limit as x approaches 0 of ln(x+e^{2x}) ?

The limit as x approaches 0 of ln(x+e^{2x}) is 0

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+e^{2x})

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

0

Solution steps

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

Plug in the value

x = 0

= ln (0 + e^{2 \cdot 0})

Simplify

ln (0 + e^{2 \cdot 0}) : 0

= 0

\int \frac{1}{x ^{2} + 6 x + 9} d x

x^{3} (x^{2} + 1)^{7}

x^{2} sin^{9} (+ x cos^{- 2} (x))

\int 3 x^{2} + 2 x - 3 d x

(4 y + t^{2} y) d y - (2 t + t y^{2}) d t = 0

What is the limit as x approaches 0 of ln(x+e^{2x}) ?
The limit as x approaches 0 of ln(x+e^{2x}) is 0