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

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

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

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

- 6

Solution steps

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= - 6

3 x

\frac{\partial}{\partial x} (x^{2} + 4 y^{2})

\int sin (x) cos (x) x d x

x \to 1 + lim (\frac{5 x + 6}{x - 1})

x = - 5, x = 1, y = 7 x, y = x^{2} - 8

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