What is the limit as x approaches infinity of (e^{x^3})/(7-x^4) ?

The limit as x approaches infinity of (e^{x^3})/(7-x^4) is -infinity

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limit as x approaches infinity of (e^{x^3})/(7-x^4)

x \to \infty lim (\frac{e ^{x^{3}}}{7 - x ^{4}})

- \infty

Solution steps

x \to \infty lim (\frac{e ^{x^{3}}}{7 - x ^{4}})

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Simplify

\frac{3 ( 27 e ^{x^{3}} x ^{8} + 108 e ^{x^{3}} x ^{5} + 60 e ^{x^{3}} x ^{2} )}{- 24} : - \frac{1}{8} (27 e^{x^{3}} x^{8} + 108 e^{x^{3}} x^{5} + 60 e^{x^{3}} x^{2})

= x \to \infty lim (- \frac{1}{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)

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

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

With the exception of indeterminate form

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

x \to \infty lim (27 e^{x^{3}} x^{8}) = \infty

x \to \infty lim (108 e^{x^{3}} x^{5}) = \infty

x \to \infty lim (60 e^{x^{3}} x^{2}) = \infty

= - \frac{1}{8} (\infty + \infty + \infty)

Simplify

- \frac{1}{8} (\infty + \infty + \infty) : - \infty

= - \infty

f^{^{'}} (x) = \frac{1}{x}

y = x^{3} - 2 x + 1

x \to 0 lim (\frac{∣ x ( h + 1 ) ∣}{x})

\int_{0}^{8} \frac{x}{x ^{2} - 1} d x

\frac{\partial}{\partial y} (x^{4} + y^{4} - 4 x y + 2)

What is the limit as x approaches infinity of (e^{x^3})/(7-x^4) ?
The limit as x approaches infinity of (e^{x^3})/(7-x^4) is -infinity