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})

Plug in the value

x = 0

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

Simplify

= - 6

f (t) = \frac{t sin ( t )}{1 + t}

\int \frac{2 - x}{x} d x

\int x^{2} d x

n = 1 \sum \infty \frac{n !}{2 n ! + 1}

x^{2} - x - 2, 3

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