What is the limit as x approaches infinity of x^3*e^{1/(x^3)}-x^3 ?

The limit as x approaches infinity of x^3*e^{1/(x^3)}-x^3 is 1

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limit as x approaches infinity of x^3*e^{1/(x^3)}-x^3

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

1

Solution steps

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

Apply the following algebraic property

: a + b = a (1 + \frac{b}{a})

x^{3} e^{\frac{1}{x ^{3}}} - x^{3} = x^{3} (e^{\frac{1}{x ^{3}}} - 1)

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: x \to \infty lim \frac{- \frac{3 e ^{\frac{1}{x ^{3}}}}{x ^{4}}}{- \frac{3}{x ^{4}}}

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

Simplify

\frac{- \frac{3 e ^{\frac{1}{x ^{3}}}}{x ^{4}}}{- \frac{3}{x ^{4}}} : e^{\frac{1}{x ^{3}}}

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

Apply the Limit Chain Rule:

1

= 1

x \to 0 + lim (e^{\frac{1}{x}} x)

x \to - 7 - lim (f (x))

x \to 1 lim (\frac{2 x - 1}{x - 1})

x \to 4 lim (x + 5)

x \to 0 lim ((1 + 5 x)^{\frac{7}{x}})

What is the limit as x approaches infinity of x^3*e^{1/(x^3)}-x^3 ?
The limit as x approaches infinity of x^3*e^{1/(x^3)}-x^3 is 1