What is the limit as t approaches-infinity of t^3e^t ?

The limit as t approaches-infinity of t^3e^t is 0

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limit as t approaches-infinity of t^3e^t

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: t \to - \infty lim (\frac{3 t ^{2}}{- e ^{- t}})

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

Apply L'Hopital's Rule

: t \to - \infty lim (\frac{6 t}{e ^{- t}})

= t \to - \infty lim (\frac{6 t}{e ^{- t}})

Apply L'Hopital's Rule

: t \to - \infty lim (\frac{6}{- e ^{- t}})

= t \to - \infty lim (\frac{6}{- e ^{- t}})

Simplify

\frac{6}{- e ^{- t}} : - 6 e^{t}

= t \to - \infty lim (- 6 e^{t})

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

= - 6 \cdot t \to - \infty lim (e^{t})

Apply the common limit:

t \to - \infty lim (e^{t}) = 0

= - 6 \cdot 0

Simplify

= 0

x \to 4 + lim (\frac{4 - x}{4 - x})

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

x \to 0 lim (\frac{8 ^{x} - 3 ^{x}}{x})

x \to 0 lim (1 - \frac{3}{x}^{x})

n \to \infty lim (\frac{- 1}{n})

What is the limit as t approaches-infinity of t^3e^t ?
The limit as t approaches-infinity of t^3e^t is 0