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

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

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

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

m

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= m

x \to \infty lim (3 x + 5)

m \to 0 lim (\frac{sin ( m )}{m})

x \to 1 - lim (\frac{arccos ( x )}{1 - x ^{2}})

x \to - 6 + lim (\frac{x}{x + 6})

x \to 2 lim (3 x^{2} + 4)

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