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

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

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

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

2

Solution steps

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

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

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= 2

3 e^{- x^{2}}

y = x^{3} sin^{2} (x)

f (x) = e^{x^{2} - x}

3 x + 1, at a = 8

\int sin (\frac{1 - x}{3}) d x

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