What is the limit as x approaches 0 of (ln(6x+1))/x ?

The limit as x approaches 0 of (ln(6x+1))/x is 6

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limit as x approaches 0 of (ln(6x+1))/x

x \to 0 lim (\frac{ln ( 6 x + 1 )}{x})

6

Solution steps

x \to 0 lim (\frac{ln ( 6 x + 1 )}{x})

Apply L'Hopital's Rule

= x \to 0 lim (\frac{\frac{6}{6 x + 1}}{1})

Plug in the value

x = 0

= \frac{\frac{6}{6 \cdot 0 + 1}}{1}

Simplify

= 6

\int \frac{r}{4 - r ^{2}} d r

\int (5 x - 1)^{3} d x

t an g e n t \frac{3 x ^{2}}{( 3 - 2 x ) ^{4}}

\int (2 t e^{t}) d t

t an g e n t y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (- 3, 2)

What is the limit as x approaches 0 of (ln(6x+1))/x ?
The limit as x approaches 0 of (ln(6x+1))/x is 6