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

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

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

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

k

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= k

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