What is the limit as x approaches 0 of (ln(3x+1))/x ?

The limit as x approaches 0 of (ln(3x+1))/x is 3

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limit as x approaches 0 of (ln(3x+1))/x

x \to 0 lim (\frac{ln ( 3 x + 1 )}{x})

3

Solution steps

x \to 0 lim (\frac{ln ( 3 x + 1 )}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{\frac{3}{3 x + 1}}{1})

= x \to 0 lim (\frac{\frac{3}{3 x + 1}}{1})

Plug in the value

x = 0

= \frac{\frac{3}{3 \cdot 0 + 1}}{1}

Simplify

= 3

19 - x^{2}

\frac{\partial}{\partial x} (ln (x^{2} + y^{2}))

y^{^{''}} = 4 y

\int tan (2 x) (sec (2 x))^{2} d x

\frac{s + 12}{( s ^{2} + 4 s )}

What is the limit as x approaches 0 of (ln(3x+1))/x ?
The limit as x approaches 0 of (ln(3x+1))/x is 3