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

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

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

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

a

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= a

x \to \frac{π}{2} lim (\frac{tan ( x )}{x})

x \to - \frac{1}{2} lim (1 - 5 x^{3})

x \to 2 lim (\frac{5 x}{4 - x ^{2}})

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x \to \infty lim (x ln (1 + \frac{3}{x}))

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