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

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

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

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

- 6

Solution steps

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

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

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

Apply L'Hopital's Rule

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

Plug in the value

h = 0

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

Simplify

= - 6

d er i v a t i v e 2 t^{3}

\int \frac{2}{cos ( 2 x )} d x

\frac{d}{d x} (\frac{x ^{2} + y - 1}{x + 1})

d er i v a t i v e f (x) = x^{10}

d er i v a t i v e x^{2} + 4 x - 7

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