What is the limit as x approaches infinity of \sqrt[3]{x^2}*e^{-x} ?

The limit as x approaches infinity of \sqrt[3]{x^2}*e^{-x} is 0

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limit as x approaches infinity of \sqrt[3]{x^2}*e^{-x}

0

Solution steps

= x \to \infty lim (e^{- x} x^{\frac{2}{3}})

Rewrite for L'Hopital

= x \to \infty lim (\frac{x ^{\frac{2}{3}}}{\frac{1}{e ^{- x}}})

Apply L'Hopital's Rule

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

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

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

x \to \infty lim (1) = 1

= \frac{2}{3} \cdot \frac{1}{\infty}

Simplify

\frac{2}{3} \cdot \frac{1}{\infty} : 0

= 0

f (x) = 8 x^{2} - x^{3}, (1, 7)

(x - 5)^{\frac{7}{3}}

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

x^{2} + 4 x - 1

\frac{\partial}{\partial x} (2 x y + 5 y^{2} - 3 x y^{3})

What is the limit as x approaches infinity of \sqrt[3]{x^2}*e^{-x} ?
The limit as x approaches infinity of \sqrt[3]{x^2}*e^{-x} is 0