What is the limit as x approaches 0 of (log_{e}(1+3x^2))/(x(e^x-1)) ?

The limit as x approaches 0 of (log_{e}(1+3x^2))/(x(e^x-1)) is 3

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 (log_{e}(1+3x^2))/(x(e^x-1))

x \to 0 lim (\frac{lo g _{e} ( 1 + 3 x ^{2} )}{x ( e ^{x} - 1 )})

3

Solution steps

x \to 0 lim (\frac{lo g _{e} ( 1 + 3 x ^{2} )}{x ( e ^{x} - 1 )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{\frac{6 x}{1 + 3 x ^{2}}}{e ^{x} - 1 + e ^{x} x})

= x \to 0 lim (\frac{\frac{6 x}{1 + 3 x ^{2}}}{e ^{x} - 1 + e ^{x} x})

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= x \to 0 lim (\frac{6 x}{( 1 + 3 x ^{2} ) ( e ^{x} - 1 + e ^{x} x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{6}{3 e ^{x} x ^{3} + 12 e ^{x} x ^{2} + 7 e ^{x} x - 6 x + 2 e ^{x}})

= x \to 0 lim (\frac{6}{3 e ^{x} x ^{3} + 12 e ^{x} x ^{2} + 7 e ^{x} x - 6 x + 2 e ^{x}})

Plug in the value

x = 0

= \frac{6}{3 e ^{0} \cdot 0 ^{3} + 12 e ^{0} \cdot 0 ^{2} + 7 e ^{0} \cdot 0 - 6 \cdot 0 + 2 e ^{0}}

Simplify

\frac{6}{3 e ^{0} \cdot 0 ^{3} + 12 e ^{0} \cdot 0 ^{2} + 7 e ^{0} \cdot 0 - 6 \cdot 0 + 2 e ^{0}} : 3

= 3

\frac{d}{d x} (sin^{2} (3 x - 2))

\frac{d}{d x} (4 arcsin (x))

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

\int 3 x^{2} d x

\frac{\partial}{\partial y} (ln (1 + x y))

What is the limit as x approaches 0 of (log_{e}(1+3x^2))/(x(e^x-1)) ?
The limit as x approaches 0 of (log_{e}(1+3x^2))/(x(e^x-1)) is 3