What is the limit as x approaches 0+of 4x^{ln(x)} ?

The limit as x approaches 0+of 4x^{ln(x)} is infinity

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limit as x approaches 0+of 4x^{ln(x)}

x \to 0 + lim (4 x^{l n (x)})

\infty

Solution steps

x \to 0 + lim (4 x^{l n (x)})

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

= 4 \cdot x \to 0 + lim (x^{l n (x)})

Apply exponent rules

= 4 \cdot x \to 0 + lim (e^{l n (x) l n (x)})

Apply the Limit Chain Rule:

\infty

= 4 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

x \to 1 lim (\frac{2 x - 2}{3 26 + x - 3})

x \to 3 lim (\frac{12 - x}{x + 3})

x \to \frac{π}{2} + lim (tan (π - x))

x \to 3 lim (2 x - 2)

x \to 5 lim (3 x + 5)

What is the limit as x approaches 0+of 4x^{ln(x)} ?
The limit as x approaches 0+of 4x^{ln(x)} is infinity