What is the limit as x approaches infinity of (x^2)/(e^{x^3)} ?

The limit as x approaches infinity of (x^2)/(e^{x^3)} is 0

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limit as x approaches infinity of (x^2)/(e^{x^3)}

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

0

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

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

= \frac{2}{3} \cdot x \to \infty lim (\frac{1}{3 e ^{x^{3}} x ^{4} + 2 e ^{x^{3}} 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

= \frac{2}{3} \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 3 e ^{x^{3}} x ^{4} + 2 e ^{x^{3}} x )}

x \to \infty lim (1) = 1

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

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

Simplify

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

= 0

\frac{d}{d x} (17 cos (x))

x \to - 2 lim (\frac{x ^{2} + 5}{x + 2})

F (T) = \frac{7}{5}

y^{^{''}} - a (y + b) = 0

f (x) = x^{3} + e^{4 x} x^{2}

What is the limit as x approaches infinity of (x^2)/(e^{x^3)} ?
The limit as x approaches infinity of (x^2)/(e^{x^3)} is 0