What is the limit as x approaches 0+of 2ln(x) ?

The limit as x approaches 0+of 2ln(x) is -infinity

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limit as x approaches 0+of 2ln(x)

x \to 0 + lim (2 ln (x))

- \infty

Solution steps

x \to 0 + lim (2 ln (x))

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

= 2 \cdot x \to 0 + lim (ln (x))

Apply the common limit:

x \to 0 + lim (ln (x)) = - \infty

= 2 (- \infty)

Apply Infinity Property:

c \cdot (- \infty) = - \infty

= - \infty

x \to 1 lim (\frac{x + 1}{x + 1})

x \to 0 - lim (\frac{x ^{4} - 1}{x ^{4}})

x \to 0 lim (e^{\frac{x - 1}{x}})

x \to 0 lim (\frac{2 x}{x ^{2} + x})

x \to \infty lim (- x^{2} e^{x})

What is the limit as x approaches 0+of 2ln(x) ?
The limit as x approaches 0+of 2ln(x) is -infinity