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

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

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

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

2

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= 2

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y^{^{'}} = x^{2} + 2 y

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