What is the limit as x approaches-infinity of-x^2e^x ?

The limit as x approaches-infinity of-x^2e^x is 0

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limit as x approaches-infinity of-x^2e^x

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

0

Solution steps

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

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

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Simplify

\frac{2}{e ^{- x}} : 2 e^{x}

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

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

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

Apply the common limit:

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

= - 2 \cdot 0

Simplify

= 0

x \to - 1 lim (2 x^{2} - 5 x - 3)

x \to 0 lim (\frac{sin ( x - π )}{x})

h \to 0 lim (\frac{e ^{h} - 1}{2 h})

h \to 0 lim (cos (h))

x \to - 1 lim (\frac{1 - x ^{4}}{x + 1})

What is the limit as x approaches-infinity of-x^2e^x ?
The limit as x approaches-infinity of-x^2e^x is 0