What is the limit as x approaches 0+of e^{-2x}ln(x) ?

The limit as x approaches 0+of e^{-2x}ln(x) is -infinity

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limit as x approaches 0+of e^{-2x}ln(x)

x \to 0 + lim (e^{- 2 x} ln (x))

- \infty

Solution steps

x \to 0 + lim (e^{- 2 x} ln (x))

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

With the exception of indeterminate form

= x \to 0 + lim (e^{- 2 x}) \cdot x \to 0 + lim (ln (x))

x \to 0 + lim (e^{- 2 x}) = 1

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

= 1 \cdot (- \infty)

Apply Infinity Property:

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

= - \infty

\frac{1}{x}, 3

\frac{1}{2} x y

6 x + \frac{16}{3}

\frac{\partial}{\partial x} (2 sin (2 x))

5 t

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