What is the limit as x approaches infinity of x^4e^{-x^3} ?

The limit as x approaches infinity of x^4e^{-x^3} is 0

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limit as x approaches infinity of x^4e^{-x^3}

x \to \infty lim (x^{4} e^{- x^{3}})

0

Solution steps

x \to \infty lim (x^{4} e^{- x^{3}})

Rewrite for L'Hopital

= x \to \infty lim (\frac{x ^{4}}{\frac{1}{e ^{- x^{3}}}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{4 x ^{3}}{e ^{x^{3}} \cdot 3 x ^{2}})

= x \to \infty lim (\frac{4 x ^{3}}{e ^{x^{3}} \cdot 3 x ^{2}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{12 x ^{2}}{3 ( 3 e ^{x^{3}} x ^{4} + 2 e ^{x^{3}} x )})

= x \to \infty lim (\frac{12 x ^{2}}{3 ( 3 e ^{x^{3}} x ^{4} + 2 e ^{x^{3}} x )})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{24 x}{3 ( 9 e ^{x^{3}} x ^{6} + 18 e ^{x^{3}} x ^{3} + 2 e ^{x^{3}} )})

= x \to \infty lim (\frac{24 x}{3 ( 9 e ^{x^{3}} x ^{6} + 18 e ^{x^{3}} x ^{3} + 2 e ^{x^{3}} )})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{24}{3 ( 27 e ^{x^{3}} x ^{8} + 108 e ^{x^{3}} x ^{5} + 60 e ^{x^{3}} x ^{2} )})

= x \to \infty lim (\frac{24}{3 ( 27 e ^{x^{3}} x ^{8} + 108 e ^{x^{3}} x ^{5} + 60 e ^{x^{3}} x ^{2} )})

Simplify

= x \to \infty lim (\frac{8}{27 e ^{x^{3}} x ^{8} + 108 e ^{x^{3}} x ^{5} + 60 e ^{x^{3}} x ^{2}})

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

= 8 \cdot x \to \infty lim (\frac{1}{27 e ^{x^{3}} x ^{8} + 108 e ^{x^{3}} x ^{5} + 60 e ^{x^{3}} x ^{2}})

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

= 8 \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 27 e ^{x^{3}} x ^{8} + 108 e ^{x^{3}} x ^{5} + 60 e ^{x^{3}} x ^{2} )}

x \to \infty lim (1) = 1

x \to \infty lim (27 e^{x^{3}} x^{8} + 108 e^{x^{3}} x^{5} + 60 e^{x^{3}} x^{2}) = \infty

= 8 \cdot \frac{1}{\infty}

Simplify

8 \cdot \frac{1}{\infty} : 0

= 0

n = 0 \sum \infty \frac{1}{n + 1}

\int \frac{x ^{2} - 16}{x ^{3}} d x

f (x) = \frac{2 ln ( x )}{x}, at x = 1

\frac{d}{d x} (\frac{sin ( 8 x )}{e ^{4 x}})

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

What is the limit as x approaches infinity of x^4e^{-x^3} ?
The limit as x approaches infinity of x^4e^{-x^3} is 0